Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Synopsis
WBBSE Class 10 Behaviour of Gases Overview
1. Pressure of a gas is defined as the normal force exerted by the gas molecules on the unit area of the wall of the container (in which the gas is kept).
2. Units of pressure in the CGS system and SI are Dyn/cm2 and N/m2, respectively. N/m2 is also called Pa (pascal). 1 Pa = 10 Dyn/cm2
3. Dimensional formula of pressure is ML–1T-2.
4. The volume of a gaseous substance means the volume of the container in which it is kept.
5. Unit of volume in SI is m3 and in the CGS system is cm3. One practical unit of volume is liter (L).
1 m3 = 1000 L , 1L = 1000 cm3,
1 m3 = 106cm3
6. Dimensional formula of volume is L3.
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7. Manometer is used to measure the pressure of a gas or air. This is also called a pressure gauge.
8. The pressure exerted by a 76 cm long column of mercury at 0°C at sea level at the latitude of 45° is called the standard atmospheric pressure or normal atmospheric pressure (atm).
1 atm=1.013 x 106 Dyn/cm2 Two other units of pressure is the bar and torr.
1 bar =106 Dyn/cm2 and 1 torr= pressure exerted by 1 mm long mercury column = 1332.8 Dyn/cm2
9. Boyle’s law: At constant temperature, the volume of a fixed mass of a gas is inversely proportional to the pressure of the gas. If a certain mass of a gas occupies a volume V at a pressure p at a constant temperature, then according to Boyle’s law, \(V \propto \frac{1}{p}\) or, pV=K, where K is a constant.
∴ At a constant temperature, if V1, V2, V3……are volumes of a fixed mass of gas at pressure p1,p2, p3 ….., respectively, then according to this law, p1V1=p2V2=p3V3=…. = constant.
10. Graphical representation of Boyle’s law:
(1)p-V graph: Plotting volume of a fixed mass of gas at a constant temperature, along) the horizontal axis and the corresponding pressure along the vertical axis, the obtained graph is a rectangular hyperbola.
(2)pV-p graph: Plotting p along the horizontal axis and corresponding pV along the vertical axis, the obtained graph is a straight line parallel to the p axis.
11. Charles’ law: At constant pressure, the volume of a given mass of a gas increases or decreases by \(\frac{1}{273}\) the part of its volume at 0°C for every 1°C rise or fall in temperature.
Let, at constant pressure volume of a fixed mass of gas at 0°C is V0.
According to Charles’ law, an increase in volume for 1°C rise in temperature= \( \frac{V_0}{273}\).
∴increase in volume for t°C rise in temperature=\(\frac{V_0 t}{273}\).
volume of the gas at t°C is Vt= \(V_0+\frac{V_0 t}{273}\) = \(V_0\left(1+\frac{t}{273}\right)\)
Similarly the volume of the gas at -t°C is V-t = \(V_0\left(1-\frac{t}{273}\right)\)
12. At constant pressure, the volume of a certain mass of gas becomes zero at -273°C. This temperature is called absolute zero. If measured accurately, the value of absolute zero temperature is -273.150C. This does not happen in reality because gas is converted into liquid much before it reaches this temperature and Charles’ law is not applicable to liquid.
13. Physicist Lord Kelvin introduced a new scale of measurement of temperature, whose zero point is taken as -273°C and the value of each degree is taken to be equal to one degree of Celsius scale. This scale is called the absolute scale or Kelvin scale of temperature.
If the temperature of a body is t°C on the Celsius scale and T K in the Kelvin scale, then T = t + 273.
14. Alternative form of Charles’ law: At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.
15. Charles’ law using absolute temperature: If a certain mass of a gas occupies a volume V at absolute temperature T at constant pressure, then according to Charles’ law,
V ∝ T or, V=KT, K = constant.
∴ if at a constant pressureV1, V2, V3… be the volume of a fixed mass of gas at – temperature T1K, T2K, T3K… respectively,
then, \(\frac{V_1}{T_1}=\frac{V_2}{T_2}=\frac{V_3}{T_3}\)=….= constant
16. Graphical representation of Charles’ law: For a certain mass of a gas at constant pressure, plotting temperature t (in°C) along the horizontal axis and volume V along the vertical axis, the graph obtained is a straight line, The straight line touches the horizontal axis at -273°C.
For a certain mass of a gas at constant pressure, plotting absolute temperature T (in K) along the horizontal axis and volume V along the vertical axis the graph obtained is a straight line passing through the origin.
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Short And Long Answer Type Questions
Understanding Gas Laws in Physics
Question 1. Explain the pressure of a gas on the pressure of the gas basis of molecular motion and the collision of gas molecules with the walls of the vessel.
Answer:
The pressure of a gas on the pressure of the gas basis of molecular motion and the collision of gas molecules with the walls of the vessel.
The force of attraction amongst the molecules of gaseous material is negligible. So the gas molecules are always moving randomly. The molecules always collide amongst themselves and also with the walls of the vessel and undergo a change of momentum.
The rate of change of momentum is force, i.e., the wall exerts force on the molecules. In reaction to the force exerted by the walls, gas molecules also exert force on the walls. This force per unit area acting perpendicularly on the wall of the vessel is the gas.
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Question 2. Define the pressure of a gas. What are the units of pressure in the CGS system and SI? Establish a relationship between these two units.
Answer:
Pressure of a gas:
1. Pressure of a gas is the force per unit area exerted by the gas molecules on the walls of the vessel.
2. Units of pressure in CGS system and SI are Dyn/cm2 and N/m2, respectively.
3. The relation between the units is \(1 \mathrm{~N} / \mathrm{m}^2=\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}=10 \mathrm{dyn} / \mathrm{cm}^2\)
Question 3. What is the volume of the gas? What are the units of volume in CGS system and SI?
Answer:
The volume of the gas:
1. The volume of a gas is the volume of the container in which it is kept.
2. The units of volume in CGS system is cm3 or ml and in SI is m3.
Question 4. The height of the mercury column is h1 in the open-end arm of the manometer U -tube and it is h2 in the other arm of the U-tube. Compare the pressure of enclosed gas in the manometer with the atmospheric pressure: (1) when h1>h2, (2) when h1<h2, (3) when h1= h2.
Answer:
Given
The height of the mercury column is h1 in the open-end arm of the manometer U -tube and it is h2 in the other arm of the U-tube.
1. When h1>h2, it is understood that the pressure of the enclosed gas is more than the atmospheric pressure. If the pressure of the enclosed gas is p1 and the atmospheric pressure of that place is p0, then p1 = p0 + Pressure of a mercury column of height (h1 – h2)
∴p1± = p0 + (h1 – h2)dg where the density of mercury is d and acceleration due to gravity at that place is g.
2. If h1<h2, it is understood that the pressure of the enclosed gas is less than atmospheric pressure. If the pressure of the enclosed gas is p2, then p2= p0 – (h2 – h1)dg.
3. If h1 = h2, it is understood that the pressure of the enclosed gas inside the manometer is equal to the atmospheric pressure. In this case, if the pressure of the enclosed gas is p3, then p3= p0.
Question 5. By taking some amount of air on the top of the mercury column in a defective barometer, determine the pressure of the atmosphere.
Answer:
Let us assume, the height of mercury in this defective barometer is H1 cm and the height of air column above the mercury column is x1 cm. If the area of cross-section inside the tube is A cm2, then the volume of air, V1 = x1 A cm3. If the atmospheric pressure is equal to the height of a mercury column of H cm, the pressure of this V1 volume of air.
p1 = pressure of (H — H1) cm of mercury column. Now the open end of the barometer immersed in mercury is raised a little upward. If the height of mercury in the tube is H2 and the height of air column above the mercury column is x2 cm then the volume of enclosed air, V2= x2A cm3. Hence, the pressure of air, p2= pressure of a mercury column of height (H – H2)cm
Since the mass of enclosed air and temperature remain constant, we get from Boyle’s law,
p1V1=p2V2
or, (H- H1).x1A = (H-H2). x2A
or, Hx1– H1x1 = Hx2– H2x2
or, H(x1-x2) =H1x1 – H2x2
∴\(H=\frac{H_1 x_1-H_2 x_2}{x_1-x_2}\)
Boyle’s Law Explained with Examples
Question 6. Write down Boyle’s law and explain it. Or, State the law which establishes the relation p1V1=p2V2. Or, Write down the law from which the relation between the volume and the pressure of an ideal gas is known.
Answer:
1. Boyle’s law: At constant temperature, the volume of a fixed mass of a gas is inversely proportional to the pressure of the gas.
Explanation: If V and p are the volumes and the pressure of the gas respectively, then according to Boyle’s law
\(V \propto \frac{1}{p}\) (at constant temperature)
or \(V=\frac{K_1}{p}\) or, pV =K1
where K1 is a constant whose value depends on the mass and the temperature of the gas. Therefore, with the temperature of a fixed mass of gas remaining constant, the product of the pressure and the volume of the gas is constant. Now, if the temperature of a fixed mass of a gas remains constant and V1, V2, and V3 are the volumes of the gas at pressures p1, p2, and p3 respectively, then according to Boyle’s law,
Charles’s Law and Its Applications
Question 7. Write down Charles’ law and explain it.
Answer:
Charles’ law: At constant pressure, the volume of a given mass of a gas increases or decreases by \(\frac{1}{273}\) of its volume at 0°C for every 1°C rise or fall in temperature
Explanation: Let the volume of a given mass of a gas at constant pressure be V0 at 0°C. By keeping the pressure constant, if the temperature of the gas is changed, then according to Charles’ law at 1°C the volume of the gas becomes
\(V_1=V_0+\frac{V_0}{273}=V_0\left(1+\frac{1}{273}\right)\)
At 2°C, the volume of the gas becomes
\(V_2=V_0+\frac{2 V_0}{273}=V_0\left(1+\frac{2}{173}\right)\)
In a similar way, at f°C the volume of the gas becomes
\(V_t=V_0\left(1+\frac{t}{273}\right)\)
Similarly, keeping the pressure constant, if the temperature is decreased by t°C, the volume of the gas becomes
\(V_t^{\prime}=V_0\left(1-\frac{t}{273}\right)\)
Question 8. Draw the graph of V-t of Charles’ law and give an idea of absolute zero temperature from it.
Answer:
At constant pressure, if a graph of a given mass of gas is drawn by taking a temperature (t) as abscissa and volume ( V) at that temperature as ordinate, then the curve obtained is a straight line.
If the straight line is extended backward, it cuts! the temperature axis at -273°C. This means that at constant pressure, the volume of a gas ’ becomes zero at -273°C. This temperature is called the absolute zero temperature.
Question 9. Write Charles’ law according to the absolute scale of temperature.
Answer:
Charles’ law according to the absolute scale of temperature:
If the volume of any gas of a given mass is V0 at 0°C temperature, then according to Charles law, the volume of the gas at constant pressure and t1°C becomes
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\(V_1=V_0\left(1+\frac{t_1}{273}\right)\) or, \(V_1=V_0 \cdot \frac{273+t_1}{273}\) or, \(V_1=\frac{T_1}{273} V_0\) ……..(1)
where T1 is the reading of t1°C temperature in absolute scale. Similarly, volume of the same gas at constant pressure and t2°C becomes
\(V_2=V_0\left(1+\frac{t_2}{273}\right)\) or, \(V_2=V_0\left(\frac{273+t_2}{273}\right)\) or, \(V_2=\frac{T_2}{273} \cdot V_0\) ………(2)
where T2 is the reading of t2°C temperature in absolute scale. Now, adding equations (1) and (2), we get
\(\frac{V_1}{V_2}=\frac{T_1}{T_2}\) or, \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) or, \(\frac{V}{T}\) = constant
∴ V∞T
So, at constant pressure, the volume of a given mass of gas is directly proportional to the absolute temperature of the gas. This is an alternative form of Charles’ law.
Question 10. Why is the term ‘fixed mass of a gas’ mentioned while stating Boyle’s law?
Answer:
‘fixed mass of a gas’:
The term ‘fixed mass of gas is always mentioned at the time of stating Boyle’s law because the pressure and volume of the gas depend on mass. If the mass is changed, pressure and volume also get changed.
Question 11. Establish a relation between m3 and L
Answer:
Relation between m3 and L
1L = 103 cm3
1m3 =106 cm3= 103• 103cm3= 103 L
Question 12. Draw the p-V curve according to Boyle’s law.
Answer:
p-V curve according to Boyle’s law:
At constant temperature, the volume (V) and the pressure (p) of a fixed mass of gas are plotted. Here, V/is plotted along the abscissa, and p is plotted along the ordinate. This is the p-V graph of Boyle’s law. The nature of the curve in the graph is a rectangular hyperbola.
Gay-Lussac’s Law in Everyday Life
Question 13. Draw the pV-p curve according to Boyle’s law.
Answer:
pV-p curve according to Boyle’s law:
we find a curve where the pressure p of a gas of fixed mass under constant temperature is plotted along the abscissa and the product (pV) of the pressure and the volume is plotted along the ordinate. This is the (pV-p) curve of Boyle’s law. Its nature is a straight line and it is parallel to the pressure axis.
Question 14. In the case of a real gas, how does the total volume of the molecules of the gas influence Boyle’s law?
Answer:
In the case of real gases, we cannot neglect the volume of the gas molecules and hence, pressure can be increased only up to a definite level at a certain temperature. For this reason, Boyle’s law cannot be accurately applied at any temperature.
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Question 15. When a balloon is inflated, both its pressure and volume increase. Is Boyle’s law violated here?
Answer:
When the balloon is inflated, the mass of air inside the balloon does not remain constant. When air is pumped into the balloon, the mass of air inside it increases. We know that there are two constants in Boyle’s law—mass of the gas and its temperature. As the mass of air does not remain constant, so Boyle’s law is not applicable here. Hence, there is no question of its violation.
Question 16. Why does the volume of an air bubble increase when it goes up from deep inside a water body?
Answer:
Pressure of water increases as one goes deeper inside a water body. So, when air bubble comes upward from the depth of a water body, the pressure of water on it decreases gradually. Now suppose the temperature of water is uniform everywhere. Thus, according to Boyle’s law, when an air bubble comes upward from the depth, its volume is inversely proportional to the pressure of water, i.e., volume increases as pressure decreases.
Question 17. What is absolute scale of temperature?
Answer:
Absolute scale of temperature:
Physicist Lord Kelvin introduced a new scale for measurement of temperature whose zero point is taken as -273°C and the value of each degree is taken to be equal to one degree of Celsius scale. In SI, a temperature of 0.01°C is considered as 273.16 K in this scale. This scale is called the absolute scale of temperature.
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Question 18.Calculate the value of absolute zero temperature from Charles’ law.
Answer:
If the volume of a given mass of gas at 0°C be V0 , then according to Charles’ law, the volume of the gas at t°C and at constant pressure becomes
\(V_t=V_0\left(1+\frac{t}{273}\right)\)
Hence at -273°C, the volume of the gas becomes
\(V_{-273}=V_0\left(1+\frac{-273}{273}\right)=0\)
Therefore, according to Charles’ law, the volume of a gas becomes zero at -273°C and at constant pressure. This temperature is called ‘absolute zero temperature’.
Question 19. Draw the V’T graph of Charles’ law. Write the concept of absolute zero temperature from the graph.
Answer:
If a curve is drawn by taking the absolute temperature T of this gas as abscissa and the volume of the gas V on the ordinate, then this curve is a straight line. If the straight line is extended backward, it touches the origin. So, according to Charles’ law, the volume of a given mass of gas becomes zero at 0 K temperature, which is not possible in reality.
Question 20. Does the volume of gas become zero in reality at an absolute zero temperature?
Answer:
According to Charles’ law, the volume of any gas becomes zero at -273°C. But this does not occur in reality because long before reaching that temperature, the gas is converted into liquid and Charles’ law is not applicable to liquid.
Question 21. Calculate the value of absolute zero temperature in Fahrenheit scale
Answer:
Absolute zero temperature in Celcius scale (C) is at -273°C. Let us assume that the reading of absolute zero temperature in Fahrenheit scale is F.
Now from the eqaution, \(\frac{C}{5}=\frac{F-32}{9}\) ,
we get \(\frac{-273}{5}=\frac{F-32}{9}\)
∴F=-459.4°F
Question 22. The manufacturing company of a tube gives information regarding the pressure at which air has to be pumped in the tube of a rubber wheel of a car. But in reality, air is pumped in at a lower pressure. Why?
Answer:
Due to friction of the wheel with the road, rubber gets heated enormously. According to Charles’ law, air inside the tube increases in volume due to this. But the tyre does not increase in volume to that extent. So, air is pumped in at lower pressure so that there is no unnecessary extra pressure on the tube.
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Very Short Answer Type Questions Choose The Correct Answer
Question 1. The volume of a definite mass of gas at 0°C is V0. What will be its volume if its temperature is raised to 100°C, keeping the pressure constant?
- 293/273 V0
- 283/273 V0
- 303/273 V0
- 373/273 V0
Answer: 4. 373/273 V0
Question 2. In Boyle’s law,
- Only mass of the gas remains constant
- Only temperature of the gas remains constant
- Mass and pressure of the gas remain constant
- Mass and temperature of the gas remain constant
Answer: 4. Mass and temperature of the gas remain constant
Question 3. The volume (V) of a definite mass of an ideal gas is plotted against its temperature (f°C) at a constant pressure. What is the temperature at which the curve intersects the temperature axis?
- 0°C
- -136.5°C
- -273°C
- 273°C
Answer: 3. -273°C
Question 4. In which scale of temperature, it is not possible for the value of temperature to be negative?
- Celsius scale
- Fahrenheit scale
- Absolute scale
- All of these
Answer: 3. Absolute scale
Question 5. The pressure and volume of a definite mass of gas are given by p and V, respectively. If pressure is increased by 25% while keeping the temperature constant, what is the new volume?
- 0.6V
- 0.751V
- 0.81V
- 0.851V
Answer: 3. 0.81V
Question 6. 1 Pa = how many Dyn/cm2?
- 1
- 10
- 100
- 1000
Answer: 2. 10
Question 7. The nature of the p-V curve according to Boyle’s law Is
- Straight line,
- Circle
- Parabola
- Rectangular hyperbola
Answer: 4. Rectangular hyperbola
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Question 8. The nature of pV-p curve according to Boyle’s law is
- Straight line
- Circle
- Rectangular hyperbola
- None of the above
Answer: 1. Straight line
Question 9. The value of absolute zero in the Fahrenheit scale is
- -452.4°F
- -462.4°F
- -459.4°F
- -463.4°F
Answer: 3. -459.4°F
Question 10. If force F acts perpendicular to the plane of a closed vessel of surface area A, then the pressure of the gas is
- p = F-A
- \(p=\frac{A}{F}\)
- \(p=\frac{F}{A}\)
- \(p=\sqrt{\frac{F}{A}}\)
Answer: 3. \(p=\frac{F}{A}\)
Question 11. Unit of pressure in SI is
- N/m2
- Bar
- Torr
- Dyn/cm2
Answer: 1. N/m2
Question 12. 1m3 = how many cm3 ?
- 103
- 104
- 105
- 106
Answer: 4. 106
Question 13. 1m3 = how many L?
- 1
- 10
- 100
- 1000
Answer: 4. 1000
Question 14. The temperature of a fixed mass of gas is changed from 0°C to 30°C at constant pressure. What is the ratio of its initial and final volume?
- 91:101
- 91:100
- 91:111
- 91:121
Answer: 1. 91:101
Question 15. The temperature of a fixed mass changed from 0°C to 30°C by heating at constant pressure. How many times is the final volume of the initial volume?
- 1.2 times
- 1.5 times
- 2 times
- 3 times
Answer: 3. 2 times
Question 16. The volume of a definite mass of gas at room temperature and a pressure of 76 cm of mercury is 1 L. What is its volume at a pressure of 38 cm of mercury if the temperature remains constant?
- 1.5L
- 2L
- 3L
- 4L
Answer: 2. 2L
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Question 17. At constant temperature, if the pressure on a definite mass of gas becomes 1/4 of its initial value then the volume increases by
- 2 times
- 3 times
- 4 times
- 5 times
Answer: 2. 3 times
Question 18. Two gases of equal mass are in thermal equilibrium. If pa,pb, and Va, Vbare their respective pressure and volumes, which of the following relation is true?
- \(p_a \neq p_b ; V_a=V_b\)
- \(p_a=p_b ; V_a=V_b\)
- \(\frac{p_a}{V_a}=\frac{p_b}{V_b}\)
- \(p_a V_a=p_b V_b\)
Answer: 4. \(p_a V_a=p_b V_b\)
Question 19. An idle gas at 27°C is heated at constant pressure so as to triple its volume. The temperature of the gas will be
- 600 K
- 900°C
- 627°C
- 900°F
Answer: 3. 627°C
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Answer In Brief
Question 1. Which instrument is used to measure the pressure of air enclosed in a container?
Answer: Manometer is used to measure the pressure of air enclosed in a container.
Question 2. Pa (pascal) is a unit of which physical quantity?
Answer: Pa is the unit of pressure.
Question 3. What is the relationship between Pa and \(\frac{\mathrm{N}}{\mathrm{m}^2}\) ?
Answer: \(1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^2\)
Question 4. While measuring pressure of a gas in a closed container by a manometer, level of mercury is lower in the open-end arm than the other arm. What does it signify?
Answer: It signifies that pressure inside the closed container is lesser than the atmospheric pressure.
Question 5. While measuring pressure of a gas in a closed container with a manometer, level of mercury is lower in the open-end arm than the other arm. What does it signify?
Answer: It signifies that pressure inside the closed container is lesser than the atmospheric pressure
Question 6. Pressure of a gas depends on which factors?
Answer: The pressure of a gas depends on its mass, volume, and temperature.
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Question 7. At constant pressure, the volume of a certain quantity of a gas depends on which factor?
Answer: The volume of a certain quantity of a gas at constant pressure depends on the temperature of the gas
Question 8. Mention the constants in Boyle’s law.
Answer:
Constants in Boyle’s law are:
1. Mass of gas and
2. Temperature
Question 9. Mention constants in Charles’ law.
Answer:
Constants in Charles’ law are:
1. Mass of gas and
2. Pressure of gas.
Question 10. What is the relationship between the Celsius scale and the absolute scale of temperature?
Answer: If the temperature of a body is t°C in the Celsius scale and T K in the absolute scale, then T=t+ 273.
Question 11. What is absolute temperature?
Answer:
Absolute temperature
Absolute temperature is the temperature of a body according to a scale where zero is taken as absolute zero.
Question 12. What is the value of freezing point of water in absolute scale?
Answer: Freezing point of water in absolute scale is 273 K.
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Question 13. What is the boiling point of water in absolute scale?
Answer: The boiling point of water in absolute scale is 373 K.
Question 14. What is the nature of the V-t graph according to Charles’ law?
Answer: The V-t graph according to Charles’ law is a straight line intersecting the x-axis at -273 K.
Question 15. What is the nature of V-T graph according to Charles’ law?
Answer: The nature of the V-Tgraph according to Charles’ law is a straight line passing through the origin.
Question 16. At which temperature, the V-T curve of Charles’ law intersects the temperature axis?
Answer: The V-T curve of Charles’ law intersects the temperature axis at 0 K.
Question 17. Is it possible to have a temperature lower than the absolute temperature?
Answer: No, it is not possible to have a temperature lower than the absolute temperature.
Question 18. What would be the volume of an ideal gas at absolute zero temperature?
Answer: The volume of an ideal gas at absolute zero temperature is zero.
Question 19. What is the value of 400 K in Celsius scale?
Answer: The value of 400 K in Celsius scale = (400- 273)°C = 127°C
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Question 20. If there is a 273 K change in temperature of a body, what is the corresponding value of this change in Celsius scale?
Answer: A change of 273 K in the temperature of the body results in a change of 273°C in Celsius scale.
Question 21. If the temperature of a body changes from 270 K to 273 K, what is its corresponding increase in Celsius scale?
Answer: Increase in temperature of the body = (273- 270)K = 3 K = 3°C (ncrease of temperature in Celsius scale).
Question 22. If the temperature of a body changes from 270 K to 273 K, what is the present temperature of the body in Celsius scale?
Answer: The present temperature of the body = 273 K =0°C.
Question 23. Does the absolute zero temperature of a gas depend on its nature, mass, volume or pressure?
Answer: No, absolute zero temperature of a gas does not depend on its nature, mass, volume or pressure.
Question 24. At low pressure or high pressure do the real gases roughly follow the equation PV=KT?
Answer: Real gases follow roughly the equation pV = KT at low pressure.
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Fill In The Blanks
Question 1. The volume of a gas increases if its temperature is increased or pressure is ______.
Answer: Decreased
Question 2. The volume of a gas decreases if its temperature is decreased or pressure is ______.
Answer: Increased.
Question 3. _________ developed an air pump to experimentally verify Boyle’s law.
Answer: Robert Hooke
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Question 4. When air bubbles rise up from the bottom surface of deep water, the volume ______.
Answer: Increases
Question 5. If the volume of a fixed temperature is V0, then its volume at t°C temperature and at constant pressure is _______.
Answer: V0(1+t/273)
Question 6. In reality, we get _________ a temperature less than the absolute zero temperature.
Answer: Do not
Question 7. _______, a scientist introduced the concept of the absolute scale of temperature.
Answer: Kelvin
Question 8. Among solid, liquid, and gaseous materials, force of attraction between the molecules is highest in ________.
Answer: Solids
Question 9. _______ law is the law relating the pressure and volume of a gas of fixed mass at a constant temperature.
Answer: Boyle’s
Question 10. _______ law is the law relating the volume and temperature of a fixed mass of gas at constant pressure.
Answer: Charles’
Question 11. Due to a lack of experimental proof, the complete idea of atoms and molecules during Avogadro’s time was _______.
Answer: Hypothetical.
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Question 12. If pressure of an enclosed air is more than the atmospheric pressure, the level of mercury in the open-end arm of a manometer is ______ than the other arm.
Answer: Higher
Question 13. If pressure of an enclosed air is less than the atmospheric pressure, the level of mercury in the open-end arm of a manometer is _______ than the other arm.
Answer: lower
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law State Whether True Or False
Question 1. A pressure gauge is a device that is used to measure the pressure of the gas.
Answer: True
Question 2. According to Boyle’s law at a constant temperature, the volume of a given mass of gas is directly proportional to the pressure exerted by the gas.
Answer: True
Question 3. The volume of a certain mass of gas becomes zero at 0°C.
Answer: False
Question 4. A change of 1°C- is equal to 1 K on the Kelvin scale.
Answer: True
Question 5. Ideal gas does not obey Boyle’s law, Charles’ law, and equation of state.
Answer: True
Wb Class 10 Physical Science
Question 6. If V1 and V2 are the volumes of a fixed mass of gas at temperatures T1 and T2 respectively, then \(\frac{V_2}{V_1}=\frac{273+t_2}{273+t_1}\).
Answer: True
Question 7. A temperature of 0°C is considered as the lowest hypothetical value of temperature.
Answer: True
Question 8. The product of pressure and the volume of a fixed mass of gas at a fixed temperature remains constant.
Answer: True
Chapter 2 Behaviour Of Gases Topic A Boyle’s Law And Charle’s Law Numerical Examples Useful Relations
1. Mathematical expression of Boyle’s law: p1V1=p2V2 where, V1 and V2 are volumes of a fixed mass of a gas at pressures p1 and p2 at constant temperature.
2. Mathematical Expression of Charles’ law: \(V_t=V_0\left(1+\frac{t}{273}\right)\)
where V0 is the volume of a fixed mass of a gas at temperature 0°C and Vt is the volume of the gas at t°C at constant pressure.
3. An alternative expression of Charles’ law ‘: \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
where V1 and V2 are the volumes of a fixed mass of gas at constant pressure at a temperature T1K, and T2K respectively.
Avogadro’s Law and Molar Volume
Question 1. The volume and the pressure of a fixed mass of gas at constant temperature is 750 ml and 80 cm Hg, respectively. What should be the pressure of the gas at the same temperature to make the volume 1000 mL?
Answer: Initial pressure of the gas (p1) = 80 cm Hg
Initial volume of the gas (V1) = 750 mL
Final volume of the gas (V2) = 1000 mL
Suppose, final pressure of the gas is p2.
Since the temperature of the gas is constant, according to Boyle’s law,
p1V1=p2V2 or, \(p_2=\frac{p_1 V_1}{V_2}\)
∴ \(p_2=\frac{80 \times 750}{1000}=60 \mathrm{~cm} \mathrm{Hg}\)
Therefore, at the same temperature, the volume of the gas becomes 1000 mL at 60 cm Hg.
Question 2. The temperature of a gas of fixed mass is 27°C. At what degree Celsius temperature, the volume of this gas becomes double, if the pressure on the gas remains unchanged?
Answer: Initial temperature of the gas (T1) = 27 + 273 = 300 K and initial volume (V1)= V.
Suppose, volume is doubled at a temperature T2 and V2 = 2V.
As the pressure of the gas is constant, according to Charles’ law,
\(\frac{V_1}{V_2}=\frac{T_1}{T_2}\) or, \(V_1 T_2=V_2 T_1\) or, \(T_2=\frac{V_2 T_1}{V_1}\).
∴ \(T_2=\frac{2 V \times 300}{V}=600 \mathrm{~K}\)
Hence, final temperature of the gas in Celsius scale = (600 — 273) = 327°C
Wb Class 10 Physical Science
Question 3. Air is present inside a glass vessel at 67°C. Keeping its pressure unchanged, the temperature of the vessel is increased. At what temperature, of air will escape?
Answer: Initial temperature of air, T1= 67 + 273 = 340 K
Suppose, initial volume = V1 and 1/3 portion escapes at temperature T2
If the final volume of air is V2, then
\(V_2=\left(1+\frac{1}{3}\right) V_1 \equiv \frac{4}{3} V_1\)
As the pressure remains constant, according to Charles’ law,
\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) or, \(V_1 T_2=V_2 T_1\) or,
\(V_1 T_2=\frac{4}{3} V_1 \times 340\) or, T2=443.33K
Therefore, final temperature in Celsius scale = 453.33-273 = 180.33°C
Question 4. A balloon of volume 100 cm3 is taken at depth of 103.3 m inside a lake. What is its new volume? Atmospheric pressure is equal to the pressure of 10,33 m of water column.
Answer: Atmospheric pressure is equal to the pressure of a water column of 10.33 m. Volume (V1) of balloon on the top surface of the lake = 100 cm3.
Initial pressure (p1) = pa, where pa is the atmospheric pressure.
Thus, pressure at the bottom of the lake, p2 =pa+ Pressure a 103.3 m water column
= \(p_1+\frac{103.3}{10.33}=11 p_a\)
Suppose, the volume of air in the balloon at the bottom of the lake = V2
Now as the temperature is uniform throughout the lake, so according to Boyle’s law,
p1V1=p2V2
pa × 100
∴ \(V_2=\frac{p_1 V_1}{p_2}=\frac{p_a \times 100}{11 p_a}\)= 9.09 cm3
Hence, the volume at the bottom of the lake becomes 9.09 cm3.
Physical Science Class 10 West Bengal Board
Question 5. Some amount of air has been put into a barometer tube of length lm and as a result, its reading has come down from 76 cm to 70 cm. What is the volume of the air at standard atmospheric pressure? Standard atmospheric pressure = 76 cm Hg and the cross sectional area of the tube = 1 cm2
Answer: Cross-sectional area of the barometer tube (A) = 1 cm2
The reading of the barometer comes down from 76 cm to 70 cm when some amount of air is put inside it.
Thus, pressure of air in the barometer tube ( p2) = 76 – 70 = 6 cm Hg and its volume (V1) = (100 – 70) x 1 = 30 cm3
If the volume of the air is V2 at standard atmospheric pressure or p2 = 76cm Hg, then according to Boyle’s law,
p1V1=p2V2 or, \(V_2=\frac{p_1 V_1}{p_2}\)
∴ V2= \(\frac{6 \times 30}{76}\) = 2.368 cm3
Question 6. The volume of a gas is 3 L at 27°C. By keeping the pressure constant, if the temperature is increased to 127 °C, its volume becomes 4 L. Calculate the value of absolute zero temperature.
Answer: Suppose, absolute zero temperature = T0 °C
Therefore, initial temperature (T1) of the gas = (27 – T0) K and its volume (V1) = 3 L and final temperature (T2) = (127 – T0) K and its volume (V2) = 4L
As the pressure is constant, so according to Charles’ law,
\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) or, \(\frac{3}{27-T_0}=\frac{4}{127-T_0}\)
Question 7. At 95 cm of Hg pressure a balloon is filled with 0.8 L air. Find the volume of the balloon if pressure is decreased to 76 cm of Hg keeping the temperature constant.
Answer: According to the question, initial pressure (p1) = 95 cm of Hg, initial volume (V1) = 0.8 L, and final pressure (p2 ) = 76 cm of Hg.
Suppose the final volume of the balloon is V2 L.
As temperature and mass of the air inside the balloon are constant, according to Boyle’s law,
p1V1=p2V2 or, \(V_2=\frac{p_1 V_1}{p_2}\)
∴ \(V_2=\frac{95 \times 0.8}{76}=1.0\)
Ideal Gas Law: Formula and Applications
Question 8. Two vessels of volume V and V’ contain air at the same temperature and their pressures are p1 and p2 respectively. They are joined by a tube of negligible volume. Find the final pressure of the system (at equilibrium).
Answer: Total volume of the system V=V’ +V” Final pressure of the system (p) = pressure due to the fist gas for the entire volume V + pressure due to the second gas for the entire volume V
∴ P=p’1+p’2
As the temperature is constant, according to Boyle’s law.
p’1V= p1V’ or, \( p_1^{\prime}=\frac{p_1 V^{\prime}}{V}\)
and p’2V= p2V” or, \(p_2^{\prime}=\frac{p_2 V^{\prime \prime}}{V}\)
∴ \(p=p_1^{\prime}+p_2^{\prime}=\frac{p_1 V^{\prime}}{V}+\frac{p_2 V^{\prime \prime}}{V}\)
= \(\frac{p_1^{\prime} V^{\prime}+p_2^{\prime} V^{\prime \prime}}{V}\)
= \(\frac{\dot{p}_1^{\prime} V^{\prime}+p_2^{\prime} V^{\prime \prime}}{V^{\prime}+V^{\prime \prime}}\)
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Synopsis
1. Combined form of Boyle’s law and Charles’ law: If the volume and pressure of a given mass of gas at absolute temperature T are V and p respectively, then the combined form of Boyle’s law and Charles’ law is
\(V \propto \frac{T}{p}\) or, \(V=\frac{K T}{p}\) or, pV=KT or, \(\frac{p V}{T}=K\)
where K is a constant,-whose value depends on the mass of the gas and also on units of volume, temperature, and pressure.
2. Avogadro’s law: At same temperature and pressure, equal volumes of all gases contain the same number of molecules. Volume in Avogadro’s law refers to the volume of the space occupied by the gas and not the volume of the molecules present inside the gas.
3. Gay-Lussac’s Law: When gasses combine to form gaseous chemical compounds or compounds, the volumes of the reacting gases and that of the product (or those of the products) are in the ratio of small whole numbers, measured under the same conditions of pressure and temperature.
4. Avogadro number: Number of molecules present in 1 mole of substances. It is denoted by nA and nA = 6.022 X 1023
5. Mathematical form of Avogadro’s law: If at a particular temperature and pressure, n moles of a gas occupies volume V, then according to Avogadro’s law,
V ∞ n or, V = kn, k = constant or, V/n = constant or, \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\)
6. Combination of Boyle’s law, Charles’ law, and Avogadro’s law: Let, at temperature T K and pressure p, n moles of a gas occupies volume V.
Now, according to Boyle’s law, \(\vee \propto \frac{1}{p}\) when n, T are constant.
According to Charles’ law, V ∝ T, when n and p are constant.
According to Avogadro’s law, V ∝ {where n, when p and T are constant.
From the above relations, we get,
[layex]V \propto \frac{n T}{p}[/latex], when p, T and n all vary.
or, \(V=\frac{R n T}{p}\) R is a constant or, pV = nRT
This equation is known as the ideal gas equation for n-gram moles of an ideal gas.
7. R is called the molar gas constant or universal gas constant. The value of R is independent of the nature of the gas.
8. Units and value of R: In the CGS system, R = 8.314 x 107 erg • mol-1 • K-1 in SI, R = 8.314 J • mol-1 • K-1
9. Dimensional formula of R: dimensional formula of R is \(M L^2 T^{-2} N^{-1} \Theta^{-1}\) [dimension of temperature =Θ and amount of substance (mole) =N]
10. Ideal gas and real gas: A gas that obeys equation of state pV = nRT under all conditions is called an ideal gas. The gases which does not obey Boyle’s law or Charles’ law or Avogadro’s law or equation of state pV = nRT are called an ideal gases. But in reality, no known gas obeys the equation pV = KT except at high temperature and low pressure, hence they are called real gas.
Physical Science Class 10 West Bengal Board
11. Basic assumptions of the kinetic theory of gases:
(1)Gases are composed of a large numbers of molecules. For a particular gas the molecules are identical but they are different for different gases.
(2)Gas molecules are assumed as point masses, sum of their volumes is negligible compared to the volume of the container.
(3)Within the container the molecules are in ceaseless, random motion in all possible directions. .
(4)During motion, the molecules collide with each other and with the walls of the container. There collisions are perfectly elastic i.e., no loss in energy during collision.
(5)There is no force of attraction or repulsion between the gas molecules. Hence the potential energy of a gas molecule is zero, the total energy is its kinetic energy.
(6)The value of molecular velocities varies from zero to infinity.
(7)If temperature of the container remain unchanged, number of molecules of unit volume anywhere inside the container is always same.
12. Reasons of deviation from ideal behavior:
1. Ideal gas molecules are point masses. But real gas molecules have finite volumes.
2. There is no force of attraction or repulsion between ideal gas molecules. But weak intermoiecular force acts between real gas molecules.
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Short And Long Answer Type Questions
Question 1. Establish the combined form of Boyle’s and Charles’ law.
Answer:
The combined form of Boyle’s and Charles’ law
Suppose, pressure and volume of a fixed mass of any gas at absolute temperature T are p and V, respectively.
According to Boyle’s law, \(V \propto \frac{1}{p}\) (when mass of gas p and T are fixed).
According to Charles’ law, V ∞ T (when mass of gas and p are fixed).
Thus, according to the law of combined variation, \(V \propto \frac{T}{p}\) (when mass of a gas is fixed but both T and p are variables)
or V=KT/P or, pV=KT
∴ \(\frac{p V}{T}=K\) ……..(1) where K is a proportional constant.
Let the initial pressure, volume and absolute temperature of a fixed mass of gas be p1, V1, T1, and the final pressure, volume, and absolute temperature be p2, V2, and T2 respectively. Then from equation (1), we get
\(\frac{p_1 V_1}{T_1}=K\) and \(\frac{p_2 V_2}{T_2}=K\)
∴ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)
This is the combined form of Boyle’s law and Charles’ law.
Question 2. Under what circumstances, a real gas may behave like an ideal gas? Or, Give the arguments in favour of considering a real gas as an Ideal gas.
Answer: The volume of a definite mass of a real gas increases steadily when the pressure on it is reduced under constant temperature. Again, the volume of a definite mass of a real gas increases steadily when the temperature of it is increased under constant pressure. As the volume of the gas increases, the mutual distance between its molecules also increases.
As a result, the value of mutual attractive force between the molecules is also reduced. For an ideal gas, it is assumed that there is no mutual attractive force between the molecules. With these arguments, a real gas behaves like an ideal gas under the lowest possible pressure or the highest possible temperature.
Physical Science Class 10 West Bengal Board
Question 3. What are the contributions of Avogadro’s hypothesis?
Answer:
Contributions of Avogadro’s hypothesis:
1. Atoms and molecules are differentiated for the first time, in Avogadro’s hypothesis.
2. Gay- Lussac’s law of gaseous volume can be explained satisfactorily with the help of Avogadro’s hypothesis.
3. With this hypothesis correlation can be established between Dalton’s atomic theory and Gay-Lussac’s law of gaseous volume.
Question 4. The molar volumes of real gases at any given temperature and pressure are more or less equal and its limiting value at STP is 22.4 L • mol-1. From the above experimental facts, deduce Avogadro’s hypothesis.
Answer:
The molar volumes of real gases at any given temperature and pressure are more or less equal and its limiting value at STP is 22.4 L • mol-1.
The value of molar volume does not depend on the nature of the gaseous material. If the temperature and the pressure remain constant, then the values of molar volumes of different real gases remain nearly equal. If the volume of nmol of gas is V, then the molar volume becomes \(\frac{V}{n}\).
As the volume of a gas changes due to change” in its temperature and pressure, so the molar volume also changes due to change of temperature and pressure. It has been experimentally proved that at standard temperature and pressure, i.e., at STP, the value of the molar volume of any gaseous material is more or less the same and its limiting value is 22.4 L or 22400 mL.
This volume is called the molar volume of a gas at STP. So, at STP, 22.4 liters of every gas contain an equal number of molecules. From this experimental result, scientist Avogadro postulated his famous theory regarding the volume and molecules of any gaseous material. This theory is known as Avogadro’s hypothesis.
Question 5. Explain how Avogadro’s law combines Gay-Lussac’s law and Dalton’s atomic theory.
Answer: It is known that hydrogen chloride (HCI) gas is formed when hydrogen (H) gas and chlorine (Cl) gas react with each other. The reaction is given by H2 + CI2 = 2HCI.
According to Gay-Lussacs’s law of gaseous volume, 1 unit volume of H has to react with 1 unit volume of Cl to produce 2 units volume of HCI.
Now, according to Avogadro’s hypothesis, 1 elementary particle of H +1 elementary particle of Cl = 2 elementary particles of HCI Meanwhile according to Dalton’s atomic theory, the smallest elementary particle of matter is atom which is indivisible.
Therefore for hydrogen or chlorine, the elementary particle is molecule rather than an atom. Atoms and molecules were first differentiated while explaining Gay-Lussac’s gaseous volume from Avogadro’s hypothesis. Hence, Avogadro’s law combined Dalton’s atomic theory and Gay-Lussac’s law of gaseous volume
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 6. Show that wet air is lighter than dry air by using the molar masses of water, nitrogen, and oxygen. Or, Give the reason for air to be lighter in rainy season than in winter.
Answer: We know, \(\text { density }=\frac{\text { mass }}{\text { volume }}\). By any means, if P volume the amount of water vapor increases in air, the mass of air decreases, and as a result, the density of air decreases. The amount of nitrogen and oxygen is high in dry air.
Molar mass of nitrogen is 28 g and the molar mass of oxygen is 32 g. Again, molar mass of water (H20) = 1×2 + 16 = 18 g
Now, suppose, some amount of dry air is taken in a closed vessel. Then, by keeping temperature and volume unchanged, 4 molecules of N2 and 1 molecule of O2 are taken out of the vessel and 5 molecules of H20 are introduced.
In this case, mass (M1) of (4 molecules of N2 + 1 molecule of O2).
=\(\left(\frac{4 \times 28}{N_A}+\frac{1 \times 32}{N_A}\right)=\frac{144}{N_A} g\)
But mass of 5 molecules of H20
M2= \(\frac{5 \times 18}{N_{\mathrm{A}}}=\frac{90}{N_{\mathrm{A}}} \mathrm{g}\)
Hence, M2<M1. Clearly, total mass of wet air inside the vessel is also reduced in this process. As a result, the density of wet air inside the vessel is also reduced. Thus, the wet air of the rainy season is lighter than dry air of winter
Question 7. Establish the ideal gas equation formed by combining Boyle’s law, Charles’ law, and Avogadro’s hypothesis.
Answer: According to Boyle’s law, \(V \propto \frac{1}{p}\) [when T and n are constant] ……..(1)
According to Charles’ law, V∝T [when p and n are constant]……(2)
According to Avogadro’s hypothesis, V∝n [when p and T are constant]…(3)
where, p = pressure of gas, V = volume of gas, T = temperature (in kelvin), n = number of moles. By combined variation, we get from equations (1],(2),(3).
\(V \propto \frac{n T}{p}\) [where p and T are variables.
or, \(V=\frac{n R T}{p}\) or, pV = nRT
Where R is the molar gas constant or universal gas constant. This equation is the ideal gas For n mole of gas.
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 8. Calculate the unit of R by the dimensional analysis of ideal gas equation.
Answer:
We get \(R=\frac{p V}{n T}\) from the ideal gas equation, pV = nRT for n mol of gases.
∴ dimensional formula of R.
= dimensional formula of pressure X dimensional formula of volume/ dimensional formula of mole number X dimensional formula of temperature
= dimensional formula of force /(dimensional formula of length)2 X (dimensional formula of length)3 / dimensional formula of mole number X dimensional formula of temperature
= dimensional formula of work/ dimensional formula of mole number X dimensional formula of temperature
So, unit of universal gas constant \(R=\frac{\text { unit of work }}{\text { unit of mole number } \times \text { unit of temperature }}\)
Therefore, unit of R in SI is j • mol-1 • K-1 and unit of R in CGS system is erg •mol-1 • K-1. Here, the temperature is absolute temperature. So, its unit is expressed in K (kelvin).
Question 9. Calculate the value of universal of constant R.
Answer:
Ideal gas equation for 1 mol of any gas at STP is p0V0 = RT0
where p0 = standard pressure, T0 = standard temperature = 273 K and V0 = volume of,1 mole of gas at STP = 22400 cm3
∴p0 = pressure of 76 cm of mercury column = 76 x 13.6 X 980 Dyn/cm2
Now, we get R=\frac{p_0 V_0}{T_0}= from p0V0 = RT0 gas or, \(R=\frac{76 \times 13.6 \times 980 \times 22400}{273}\)
∴ R = 8.31 X 107 erg mol-1 • K-1 = 8.31 J. mol-1 K-1
Question 10. Write down the basic assumptions of kinetic theory of an ideal gas.
Answer: The basic assumptions of the kinetic theory of an ideal gas are as follows:
1. All the gases are composed of many molecules. Molecules of the same gas are of same nature but molecules of different gases are of different nature.
2. Gas molecules behave like point masses. So, volume of the molecules is negligible compared to the volume of the vessel.
3. Until the molecules collide amongst themselves and with the walls of the vessel, they move in straight lines with uniform speed. As a result of the collision, the motion of the molecules become disorderly.
4. The total linear momentum and the total kinetic energy of the molecules before and after the coliision remain unchanged.
5. No attractive or repulsive force acts within the gas molecules, i.e., there is no potential energy of the molecules and the entire energy is kinetic energy.
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 11. Give two arguments in favour of the mobility of gas molecules.
Answer:
1. Pressure and diffusion are the two properties of gases which prove that the gas molecules are mobile. Gas molecules always collide amongst themselves and also with the walls of the vessel.
2. A force is exerted on the wall when the molecules collide against it. Pressure of a gas is equal to this force per unit area acting perpendicular to the wall of the vessel.
Further, two or more gases mix amongst themselves on their own. This is called diffusion. Therefore, both pressure of a gas and diffusion arise due to movement of the gas molecules.
Question 12. Discuss the influence of pressure on the volume of a gas.
Answer:
1. If the pressure on an enclosed gas is increased, distance between the molecules decreases. As a result, volume of the gas decreases. Due to decrease in the volume, number of collisions per second on unit area of the wall increases.
2. As a result, pressure of the gas increases and the external pressure and the internal pressure become equal and act in opposite directions.
3. In the same way, if the external pressure is decreased, distances between the molecules increase and as a result, volume increases. Due to increase of the volume, number of collisions per second on unit area of the wall decreases and consequently, the pressure of the gas decreases.
4. Here also, the external pressure and the internal pressure become equal and act in opposite directions.
Question 13. What is diffusion? Which properties of the gas molecules are known due to diffusion? Or, If an incense stick is ignited inside the room, the fragrance of the stick spreads within the room in a very short time. This phenomenon takes place due to which property of the gas? Write down the reasons.
Answer:
1. Diffusion is the phenomenon by which two or more non-reacting gases (light or heavy) are mixed spontaneously to form a homogeneous mixture, when they come in contact with each other.
2. If an incense stick is ignited inside a room, the fragrance of the stick spreads within the room in a very short time. As soon as the stick is ignited, there is collision among the gas molecules and also with the molecules of the air. Due to frequent collisions, the direction of motion of the gas molecules gets changed every moment.
3. The intermolecular space is much more than the volume of the gas molecules. As a result, the motion of the molecules of a particular gas enables them to enter into another gas through that intermolecular space and spread within a short time.
4. So, it is known through the process of diffusion that gas molecules are in motion and their motion is disorderly.
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 14. Why the volume of a real gas of a definite mass at constant temperature and high pressure does not decrease in accordance with Boyle’s law, even if the pressure on it is increased? Or, Boyle’s law is applicable for an ideal gas but not for a real gas at high pressure. Why?
Answer:
1. According to the conditions of kinetic theory of gases, total volume of the molecules of a gas is negligible compared to the volume occupied by the gas.
2. But molecules of a real gas, though small, have a finite volume. If pressure is increased, the above mentioned condition remains valid upto a pressure of 1 atmosphere, but at higher pressure the molecules of the gas come very near to each other.
3. As a result, when the pressure is increased on the real gases already kept at high pressure, their volumes do not decrease likewise according to Boyle’s law.
Question 15. How is the presence of intermolecular forces responsible for deviation of a real gas from an ideal gas?
Answer: The pressure of a real gas is less than that of an ideal gas at same temperature and for same mass and volume. This is a deviation of an ideal gas from a real mass.
Explanation: When a gas molecule remains inside the vessel and is slightly away from the wall of the vessel, then it experiences an equal attractive force by the molecules all around it from every direction. As a result, resultant force on that molecule becomes zero. But when a molecule is very near the wall of the vessel, then the molecules inside the vessel and away from the wall exert an inward resultant force on that molecule.
As a result, the velocity of the molecule is reduced and thus strikes the wall at a lower speed. Now according to the kinetic theory of an ideal gas, there are no intermolecular forces between the molecules. So, real gas exerts comparatively less amount of pressure on the wall due to the presence of intermolecular forces.
Question 16. Mention two observations which prove the presence of mutual attractive forces between the gas molecules.
Answer: The following two observations prove the presence of intermolecular forces in a real gas:
1. Any gas gets condensed into liquid at a definite temperature when its temperature is reduced. Explanation: When temperature is reduced, kinetic energy of the gas molecules is also reduced. As a result, the molecules come nearer to each other only due to mutual attractive force. More the molecules come nearer to each other, more the value of intermolecular force increases. Hence, the gas gets condensed into liquid.
2. The pressure of a real gas is slightly less than the pressure of an ideal gas under similar conditions.
Explanation: When a molecule comes very near to the wall of a closed vessel, then that molecule exerts comparatively less force on the wall. This is due to the fact that the molecule which has come closer to the wall of the vessel is attracted by the other molecules away from the wall. So, pressure of the gas is slightly less than the pressure of an ideal gas under similar conditions.
Question 17. Determine the dimensional formula for universal gas constant.
Answer: The ideal gas equation for n mol of gases is pV = nRT; where p = pressure, V = volume, T = absolute temperature and R is the universal gas constant.
∴\(R=\frac{p V}{n T}\)
Hence, the dimensional formula of R = =\(\frac{M L^{-1} T^{-2} \times L^3}{N \times \Theta}=M L^2 T^{-2} N^{-1} \Theta^{-1}\)
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Very Short Answer Type Question Choose The Correct Answer
Question 1. Real gases behave like ideal gas at
- High pressure and low temperature
- Low pressure and high temperature
- Low pressure and low temperature
- High pressure and high temperature
Answer: 2. Low pressure and high temperature
Question 2. Molar mass of water is
- 16g
- 18g
- 20g
- 22g
Answer: 2. 18g
Question 3. Mass of 5 mol hydrogen gas is
- 5g
- 10g
- 20g
- 15g
Answer: 2. 10g
Question 4. During the collision of molecules of an ideal gas
- Only linear momentum remains conserved
- Only kinetic energy remains conserved
- Both linear momentum and kinetic energy remain conserved
- Neither linear momentum nor kinetic energy remains conserved
Answer: 3. Both linear momentum and kinetic energy remain conserved
Question 5. The energy of the molecules of an ideal gas is
- Only potential energy
- Only kinetic energy
- The sum of potential energy and kinetic energy
- The difference of kinetic energy and potential energy
Answer: 2. Only kinetic energy
Question 6. Which quantity is a constant in the equation pV = nRT?
- p
- V
- T
- R
Answer: 4. R
Question 7. Value of the universal gas constant in the CGS system is
- 8.31 × 106 erg. mol-1 K-1
- 8.31 × 107 erg . mol-1 K-1
- 8.31 × 108 erg . mol-1 K-1
- 8.31 × 109 erg . mol-1 K-1
Answer: 2. 8.31 × 107 erg . mol-1 K-1
Question 8. If x number of molecules are present in volume V of He gas at pressure p and temperature T, then the number of molecules present in a volume 3 V of O2 gas at the same pressure and temperature is
- x
- 3x
- x/3
- 9x
Answer: 2. 3x
Question 9. The freezing point of water in an absolute scale is
- 0 K
- 273 K
- 373 K
- 173 K
Answer: 2. 273 K
Question 10. The dimensional formula of universal gas constant is
- \(\mathrm{ML}^2 \mathrm{TN}^{-1} \Theta^{-1}\)
- \(M L^2 T^{-2} N^{-1} \Theta^{-1}\)
- \(M L^2 T^{-2} N^{-1} \Theta\)
- \(M L^2 T^{-1} N^{-1} \Theta^{-1}\)
Answer: 3. \(M L^2 T^{-2} N^{-1} \Theta\)
Question 11. According to the kinetic theory of gases, at what temperature the molecule of an ideal gas has zero kinetic energy?
- 273 K
- 300 K
- 0 K
- 100 K
Answer: 3. 0 K
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 12. 84g N2 = how many moles?
- 1
- 2
- 3
- 4
Answer: 3. 3
Question 13. Among 2g H2, 28g N2, and 44 g CO2, which one has the largest volume at STP?
- 2gH2
- 28gN2
- 44gCO2
- All have the same volume
Answer: 4. All have the same volume
Question 14. If Avogadro number is N, then one mol of C02 contains how many molecules?
- N
- 2N
- 3N
- 4N
Answer: 3. 3N
Question 15. The mass of one molecule of oxygen is
- 5.31 × 10-23g
- 10.62 × 10-23g
- 15.93 × 10-23g
- 2.65 × 10-23g
Answer: 1. 5.31 × 10-23g
Question 16. What is the molecular mass of CO2 when the mass of 11.2 L of CO2 at STP is 22g?
- 22
- 44
- 88
- 11
Answer: 2. 44
Question 17. The number of molecules present in 7g of N2 gas is
- 12.044 × 1023
- 1.506 × 1023
- 3.011 × 1023
- 6.022 × 1023
Answer: 2. 1.506 × 1023
Question 18. The amount of work done by expansion of 1 mol ideal gas against fixed pressure when its temperature is increased by 1K is
- R
- 2 R
- 3 R
- 4 R
Answer: 1. R
Wbbse Class 10 Physical Science Chapter 6 Question Answer
Question 19. The volume of 2.2 g of CO2 at 27°C and a pressure of 570 mm of the mercury column is
- 4.92 L
- 0.82 L
- 3.28 L
- 1.64 L
Answer: 4. 1.64 L
Question 20. The behavior of real gases show maximum deviation from the behavior of ideal gas under
- Low pressure and low temperature
- Low pressure and high temperature
- High pressure and low temperature
- High pressure and high temperature
Answer: 3. High pressure and low temperature
Question 21. There is N number of molecules in nmol of a gas. Value of – is n
- 9.033 x 1023
- 12.044 x 1023
- 3.011 x 1023
- 6.022 x 1023
Answer: 4. 6.022 x 1023
Question 22. The equation of state of 7 g of oxygen is
- pV=7RT
- \(p V=\frac{32}{7} R T\)
- \(p V=\frac{7}{32} R T\)
- pV= 14RT
Answer: 4. pV= 14RT
Question 23. What is the value of pV for 1.12 L for an ideal gas at STP?
- 2RT
- RT
- 0.05RT
- 1.12RT
Answer: 3. 0.05RT
Question 24. No of moles present per unit volume (L) of an ideal gas is
- pRT
- \(\frac{p}{R T}\)
- \(\frac{R T}{p}\)
- \(\frac{p T}{R}\)
Answer: 2. \(\frac{p}{R T}\)
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Answer In Brief
Question 1. At low temperature or high temperature do the real gases roughly follow the equation pV=KT?
Answer: Real gases roughly follow the equation pV = KT at high temperature.
Question 2. How is the pressure of a definite mass of gas related to the motion of its molecules?
Answer: Keeping the volume of a definite mass of gas constant, if the velocity of the molecules of the gas increases, pressure of the gas inside the vessel also increases.
Question 3. What is molar volume?
Answer:
Molar volume
The volume occupied by one gram mole of any gaseous substance (elementary or compound) at a fixed temperature and pressure is called the gram molar volume or molar volume.
Question 4. Does the value of molar volume depend on the nature of the gas?
Answer: No, the value of molar volume does not depend on the nature of the gas.
Question 5. The value of molar volume depends on which factor?
Answer: The value of molar volume depends on the pressure and temperature of the gas.
Question 6. What is the limiting value of molar volume of any real gas at STP?
Answer: The limiting value of molar volume of any gaseous material at STP is 22.4 L or 22400 mL.
Question 7. What is meant by the volume of a gas in Avogadro’s hypothesis?
Answer: Volume of a gas in Avogadro’s hypothesis means the volume of the space occupied by the gas.
Question 8. Who is the first scientist to differentiate between atoms and molecules?
Answer: Avogadro is the first scientist to differentiate between atoms and molecules.
Question 9. How to reduce the density of air of a place at a definite temperature?
Answer: If the amount of water vapour at a place at a definite temperature increases due to some reason, density of air gets reduced.
Question 10. The number of molecules in a balloon filled with hydrogen gas is N. What is the number of molecules in a balloon filled with an equal volume of nitrogen at the same temperature and pressure?
Answer: The number of molecules in a balloon containing equal volume of nitrogen gas at the same temperature and pressure is also N.
Question 11. Between dry air and wet air, which one has less density?
Answer: Wet air has less density than dry air.
Question 12. Is there any change in the kinetic energy of the gas molecules in a perfectly elastic collision?
Answer: In a perfectly elastic collision, kinetic energy of the gas molecules remains conserved.
Question 13. The fragrant fumes of a burning incense stick placed at one corner of the room are perceived by smell at another corner of the room. Which property of the gas molecules is demonstrated by this?
Answer: The property of motion of gas molecules is demonstrated by this.
Question 14. The fragrant fumes of a burning incense stick placed at one corner of the room are perceived by smell at another corner of the room. What information about the motion of the gas molecules is available from this phenomenon?
Answer: Due to the mutual collisions of the gas molecules, there is a continuous change in their directions—this information is available from the given phenomenon.
Question 15. What is the number of molecules in 22.4 L of a gas at STP?
Answer: The number of molecules in 22.4 L of a gas at STP is 6.022 x1023.
Question 16. How can the pressure of a certain quantity of a gas kept in a closed vessel and attached with a piston be increased without changing its temperature?
Answer: The pressure of gas inside the closed vessel increases if the piston is pushed inside the vessel, without changing the temperature.
Question 17. What is the difference between an ideal gas and a real gas on the basis of inter-molecular forces?
Answer: Though the value of intermolecular forces in the case of an ideal gas is zero, it is not so in the case of a real gas.
Question 18. Write down the ideal gas equation for n mol of a gas.
Answer: The ideal gas equation for n moles of a gas is pV=nRT.
Question 19. Write down the ideal gas equation for 1 mol of a gas.
Answer: The ideal gas equation for one mole of a gas is pV=RT.
Question 20. Is the equation pV = nRT applicable in the same way to all ideal gases?
Answer: Yes, the equation pV=nRT is applicable in the same way to all ideal gases.
Question 21. What is the unit of universal gas constant R in CGS system and SI?
Answer: The unit of universal gas constant R in CGS system and SI are erg. mol-1 . K-1 and J • mol-1. K-1, respectively.
Word Problems on Gas Laws with Solutions
Question 22. What is the value of universal gas constant R in cal . mol-1. K-1?
Answer: The value of universal gas constant in cal . mol-1 . K-1 is 1.987.
Question 23. What is the value of universal gas constant R in L . atm . mol-1. K-1 ?
Answer: The value of universal gas constant in L . atm . mol-1. K-1 is 0.082.
Question 24. What is the relation between the volume of the gas and the total volume of the molecules in case of an ideal gas?
Answer: In case of an ideal gas, total volume of the molecules of the gas is considered to be negligible as compared to the volume occupied by the gas.
Question 25. What is the nature of the collision between two molecules of an ideal gas?
Answer: The collision between two molecules of an ideal gas is an elastic collision because both linear momentum and kinetic energy are conserved in this case.
Question 26. What is the nature of the energy of gas molecules?
Answer: The gas molecules possess kinetic energy.
Question 27. What is condition of diffusion?
Answer:
Condition of diffusion
According to the condition of diffusion, when two or more gases which do not react with each other are kept in contact, they may undergo diffusion.
Question 28. Can diffusion take place against gravitation?
Answer: Yes, diffusion can take place against gravitation.
Question 29. A real gas can be converted into liquid by reducing its temperature. What can be inferred about intermolecular force from this phenomenon?
Answer: This phenomenon gives a preliminary idea about the intermolecular force in a real gas.
Question 30. Due to which condition of kinetics, the pressure of an ideal gas is greater than the pressure of a real gas under the same condition?
Answer: The condition that there is no attraction or repulsion amongst the molecules of an ideal gas is responsible for the given occurrence.
Question 31. What is the net attractive force on the molecules of a gas when it remains inside a vessel?
Answer: The net attractive force on the molecules of the gas is zero when it remains inside a vessel.
Question 32. When does a net attractive force act on the molecules of a gas inside a vessel?
Answer: A net attractive force acts on the molecules of a gas inside a vessel when they are very close to the wall of the vessel.
Question 33. Is it possible to convert an ideal gas into a liquid?
Answer: As there is no attractive force amongst the molecules of the ideal gas, hence it is not possible to convert an ideal gas into a liquid.
Question 34. Is it possible to convert a real gas into a liquid?
Answer: As attractive forces are present amongst the molecules of the real gas, it is possible to convert a real gas into a liquid.
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Fill In The Blanks
Question 1. There is _______ force of attraction between the molecules of an ideal gas.
Answer: No
Question 2. If the pressure of a definite mass of gas is quadrupled at constant temperature, ________ becomes one-fourth.
Answer: Volume
Question 3. The mass of 2 mol of CO2 at STP is __________.
Answer: 88g
Question 4. The volume of 4 g of H2 at STP is ________.
Answer: 44.8 L
Question 5. The molar mass of water is ________.
Answer: 18 g. mol-1
Question 6. The value of molar volume does not depend on _________ of the gas.
Answer: Nature
Question 7. If the number of molecules present in 2 g of H2 at STP is N, then the number of molecules present in 64 g of O2 at STP is_______.
Answer: 2N.
Question 8. If the Avogadro number is N, then number of molecules present in 8.5g of NH3 gas is _________.
Answer: N/2
Question 9. If the temperature of an enclosed gas increases, then kinetic energy of the molecules also ________.
Answer: Increase
Question 10. Compared to the time required by a gas molecule to traverse its free path, the time of collision is ________.
Answer: Neglible
Question 11. During collision of two molecules, both linear momentum and ___________ are conserved.
Answer: Kinetic Energy
Question 12. Force of attraction _________ between molecules of a real gas.
Answer: Exists
Question 13. The value of for one mole of an ideal gas is_________.
Answer: R
Question 14. Moist air is __________ than dry air.
Answer: Lighter
Question 15. In the equation \(p V=\frac{W}{M} R T\), M is the ___________.
Answer: Molar Mass
Question 16. Unit of molar mass in CGS system is _______.
Answer: g. mol-1
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law State Whether True Or False
Question 1. One mole of any gas at STP occupies 22.4 L volume and 6.022 x 1023 molecules.
Answer: True
Question 2. The value of molar volume depends upon the nature of the gas.
Answer: False
Question 3. Real gases behave like ideal gases at low pressure and high temperature.
Answer: True
Question 4. Mean free path of a gas molecule is inversely proportional to the number of molecules per unit volume.
Answer: True
Question 5. The collision between two molecules of an ideal gas is an inelastic collision.
Answer: True
Question 6. The value of universal gas constant in L atm. mol-1 . K-1 is 0.082.
Answer: True
Question 7. Dimensional formula for universal gas constant is \(M L^2 T^2 N \Theta^{-1}\).
Answer: True
Question 8. In case of an ideal gas, volume of the gas molecules is neglected.
Answer: True
Question 9. At constant pressure, the density of a gas is inversely proportional to its absolute temperature.
Answer: True
Question 10. The value of R in the equation pV=RT depends on the values of p, V, and T.
Answer: True
Question 11. The total kinetic energy of the molecules gas at 0 K is zero.
Answer: True
Chapter 2 Behaviour Of Gases Topic B Combination Of Boyle’s Law And Charles’ Law, Ideal Gas Equation, Avogadro’s Law Numerical Examples Useful Relations
1. Combination of Boyle’s law and Charles’ law: \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)
where V1, p1 and T1(K) are respectively initial volume, pressure and temperature, and V2, p2 and T2(K) are respectively final volume, pressure and temperature T2(K) of a gas of fixed mass
2. Avagadro’s law: \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\)
where, under the condition of same temperature and pressure n1 moles of a gas occupies volume V1 and n2 volume of the gas occupies the volume V2.
3. Equation of state for n moles of an ideal gas:
pV=nRT, where p,V,T and R are pressure, volume, absolute temperature and universal gas constant respectively.
4. At STP molar volume of any gas is 22.4 L or 22400 mL.
5. Avogadro number: NA = 6.022 x 1023
Question 1. Find the number of molecules in 22 g of CO2 and 9 g of water.
Answer: Avogadro number, N = 6.022 x 1023
Molar mass of C02 = 44 g
∴22 g of CO2 = mol CO2 = mol CO2
So, the number of molecules in 22 g of CO2 = \(\frac{N}{2}=\frac{6.022 \times 10^{23}}{2}=3.011 \times 10^{23}\)
Again, molar mass of H2O = 1/2 mol H2O
∴ 9g H2O = 9/18 mol H2O= 1/2 mol H2O
Hence, number of molecules in 9 g of H2O = \(\frac{N}{2}=\frac{6.022 \times 10^{23}}{2}\) = 3.011 x 1023
Question 2. What is the molar mass of N2 if the mass of 5.6 L of N2 at STP is 7 g?
Answer: Mass of 5.6 L of N2 gas at STP = 7g
∴ mass of 1L of N2 gas at STP = \(\frac{7}{5.6}\)g and mass of 22.4 L of N2 gas at STP = \(\frac{7}{5.6} \times 22.4\) = 28 g
So, molar mass of N2 = 28 g .
Question 3. what is the mass of 3 mol of NH3 gas? What is the volume of that quantity of gas at STP?
Answer: Molar mass of NH3 = (14 +1 x 3)g = 17g
∴ mass of 3 mol of NH3 gas = 17 x 3g = 51g
The volume of 3 mol of NH3, gas at STP = 22.4 X3L = 67.2 L
Question 4. How many molecules are present in 1g of N2 gas?
Answer: Molar mass of N2 = 28 g
Avogadro number, N = 6.022 x 1023
Now, there are 6.022 x 1023 molecules present in 28 g of N2 gas.
Hence, number of molecules in 1g of N2 gas \(\frac{6.022 \times 10^{23}}{28}=2: 15 \times 10^{22}\)
Question 5. Between 2 mol of N2 and 1 mol of NH3, which one has more atoms?
Answer: Number of atoms in one molecule of N2 = 2
Thus, number of atoms in 2 mol of N2 gas = 2 X 6.022 x 1023 X 2 = 24.088 x 1023
Number of atoms is one molecule of NH3 = 4
Thus, the number of atoms in 1 mol of NH3 gas = 4 X 6.022 X 1023= 24.088 x 1023
Hence, both 2 mol of N2 and 1 mol of NH3 contain the same number of atoms.
Question 6. The volume of a gas is 256 cm3 at 30°C temperature and a pressure of 108 cm Hg. What Is the volume of the gas at 0°C temperature and 76 cm Hg pressure?
Answer: Initial pressure of the gas (p1)= 108 cm
Initial temperature (T1) = 30 + 273 = 303 K
Initial volume (V1) = 256 cm3
Final pressure (p2) = pressure of 76 cm Hg
Final temperature (T2) = 0 + 273 = 273 K
Now, suppose the final volume of the gas = V2
so, \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\) or, \(\frac{108 \times 256}{303}=\frac{76 V_2}{273}\) or, V2 = 327.77 cm3
Therefore, final volume of the gas, V2 = 327.77 cm3
Question 7. The temperature of a gas of a certain mass is 27°C. The gas is heated in such a way that its pressure and volume are doubled. Calculate the final temperature of the gas.
Answer: Initial temperature (T1) of the gas = 27 + 273 = 300 K
If the initial pressure of gas, p1 = p, and initial volume V1 = V, then the final pressure, p2 = 2p and final volume, V2 = 2 V.
Suppose, the final temperature of the gas is T2 Thus by combining Charles’ law and Boyle’s law, we get
\(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\) or, \(\frac{p V}{300}=\frac{2 p \times 2 V}{T_2}\)
Hence, the final temperature of the gas in Celsius scale = (1200- 273)°C = 927°C
Question 8. The volume of a gas at 5TP is 10 L. Calculate the number of moles of the gas
Answer: Pressure of gas (p) = 1 atmosphere
Volume (V) = 10 L
Temperature (T) = 273 K
Universal gas constant (R) = 0.08205 L . atm. mol-1 . K-1
If n is the number of moles of the gas, then we get from ideal gas equation, pV= nRT or, n=pV/RT
or, \(n=\frac{1 \times 10}{0.08205 \times 273}\) = 0.446
An alternative method:
The volume of 1 mol of gas at STP = 22.4 L
Hence, number of moles of the gas,
n = 10/22.4 = 0.446
Question 9. Find the mass of 8.31 L of methane gas at 127°C and 5 atmospheric pressure. Given, molar mass of methane gas = 16 g. mol-1, density of mercury = 13.6 g/cm3
Answer: volume of methane gas (V) = 8.31L = 8.31 × 1000 cm3
Temperature (T) = 127 + 273 = 400K
Pressure (p) = 5 atm = 5 × 76 × 13.6 × 980 Dyn/cm3
Molar mass (M) = 16 g . mol-1
Let us assume that the mass of methane gas = W
From ideal gas equation, \(p V=\frac{W}{M} R T\) or, \(W=\frac{p V M}{R T}\)
or, \( W=\frac{5 \times 76 \times 13.6 \times 980 \times 8.31 \times 1000 \times 16}{8.31 \times 10^7 \times 400}\) = 20.26g (approx).
Chapter 2 Behaviour Of Gases Miscellaneous Type questions Match The Columns
Question 1. Different Curves are given in column A. Match the curves with their respective natures given in column B.
| Column A |
Column B |
| pV-p |
1. Straight line passing through the origin |
| p-V |
2. Straight line passing parallel to the pressure axis |
| V-t |
3. Straight line not passing through the origin |
| V-T |
4. Rectangular hyperbola |
Answer:
pV-p: 2. Straight line passing parallel to the pressure axis
p-V: 4. Rectangular hyperbola
V-t: 1. Straight line passing through the origin
V-T: 3. Straight line not passing through the origin
| Column A |
Column B |
| Unit of rate of diffusion |
1. 105 N . m-2. |
| 1 pa = ______ |
2. erg. mol1 . K-1 |
| Unit of the universal gas constant |
3. 10 Dyn. cm-2 |
| 1 bar = _______ |
4. m3. s-1 |
Answer:
Unit of rate of diffusion: 4. m3. s-1
1 pa = ______: 3. 10 Dyn. cm-2
Unit of the universal gas constant: 2. erg. mol1 . K-1
1 bar = _______: 1. 105 N . m-2.
