The state of a fixed mass of gas is therefore determined by the parameters<\/strong><\/p>\n The rules, that govern the change of one parameter with the change of another keeping the third one constant, are called gas laws.<\/p>\n The relationship between the volume and temperature of a fixed mass of gas at constant pressure was investigated experimentally by the French scientist Charles in 1787.<\/p>\n Statement:<\/strong> When pressure is kept constant, the volume of a fixed mass of gas increases or decreases by 1\/273 th part of its volume at 0\u00b0C, for each degree Celsius rise or fall in temperature.<\/p>\n Mathematical expression:<\/strong> Let V0<\/sub> be the volume of a mass m of a gas at 0\u00b0C. As per Charles\u2019 law, the volume at 1\u00b0C will be<\/p>\n \u21d2 \\(V_1=V_0+\\frac{V_0}{273}=V_0\\left(1+\\frac{1}{273}\\right)\\)<\/p>\n The volume of the gas at \\(2^{\\circ} \\mathrm{C}\\),<\/p>\n \u21d2 \\(V_2=V_0+\\frac{2 V_0}{273}=V_0\\left(1+\\frac{2}{273}\\right)\\)<\/p>\n \u2234 The volume of the gas at \\(t^{\\circ} \\mathrm{C}\\),<\/p>\n \u21d2 \\(V_t=V_0+\\frac{V_0 t}{273}=V_0\\left(1+\\frac{t}{273}\\right)\\)….(1)<\/p>\n Similarly, if temperature is decreased by t\u00b0C, i.e., at a temperature -t\u00b0C, the volume of the gas becomes \\(V_{-t}=V_0\\left(1-\\frac{t}{273}\\right)\\)<\/p>\n Therefore, at constant pressure volume of a fixed mass of gas changes linearly with its temperature. So, at fixed pressure, a graph plotted between the volume of a gas of a fixed mass and its temperature gives a straight line (AB).<\/p>\n The law that relates the change in pressure of a fixed mass of a gas at a fixed volume, with change in temperature is called pressure law or Regnault\u2019s law.<\/p>\n Statement<\/strong>: When volume is kept constant, the pressure of a fixed mass of gas increases or decreases by1\/273th part of its pressure at 0\u00b0C, for each degree centigrade rise or fall in temperature.<\/p>\n Mathematical expression:<\/strong> Let p0<\/sub> be the pressure of a fixed mass of a gas at 0\u00b0C. The pressure is pt<\/sub> when the temperature is raised to t \u00b0C.<\/p>\n Therefore the pressure of the gas at 1\u00b0C, \\(p_1=p_0+\\frac{p_0}{273}=p_0,\\left(1+\\frac{1}{273}\\right)\\)<\/p>\n and the pressure of the gas at t\u00b0C, \\(p_t=p_0+\\frac{p_0 t}{273}=p_0\\left(1+\\frac{t}{273}\\right)\\)<\/p>\n Similarly, the pressure of the gas at -t\u00b0C, \\(p_{-t}=p_0-\\frac{p_0 t}{273}=p_0\\left(1-\\frac{t}{273}\\right)\\)<\/p>\n Hence the change in pressure of a fixed mass of gas is linearly related to the change in temperature at constant volume.<\/p>\n So, at constant volume, a graph plotted between the pressure of a gas of fixed mass and its temperature gives a straight line.<\/p>\n The increase in either the volume or the pressure of a gas with a rise in temperature is loosely termed as thermal expansion, although an increase in pressure is not actually an expansion.<\/p>\n Charles’ law:<\/strong> Suppose a fixed mass of a gas. at constant pressure, has volume V0<\/sub> at 0\u00b0C, at V1<\/sub> at t1<\/sub>\u00b0C and V2<\/sub> at t2<\/sub>\u00b0C. From Charles\u2019 law.<\/p>\n \u21d2 \\(V_1=V_0\\left(1+\\frac{t_1}{273}\\right)=V_0\\left(\\frac{273+t_1}{273}\\right)=\\frac{V_0}{273} \\cdot T_1\\)<\/p>\n [where T1<\/sub> = t1<\/sub> + 273]<\/p>\n Obviously, the temperature t1<\/sub> in the Celsius scale is the same as the temperature T1<\/sub> K in the Kelvin scale.<\/p>\n Similarly, \\(V_2=\\frac{V_0}{273} \\cdot T_2\\)<\/p>\n where \\(T_2=t_2+273\\)<\/p>\n Here, \\(t_2{ }^{\\circ} \\mathrm{C}=T_2 \\mathrm{~K}\\).<\/p>\n \u2234 \\(\\frac{V_1}{V_2}=\\frac{T_1}{T_2}=\\) constant<\/p>\n or, \\(V \\propto T\\) at constant pressure.<\/p>\n Hence, Charles’ law can also be stated as the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in an absolute scale.<\/p>\n Pressure law:<\/strong> Suppose a fixed mass of a gas at constant volume has pressure p0<\/sub> at 0\u00b0C, p1<\/sub> at t1 <\/sub>\u00b0C. and p2<\/sub> at t2 <\/sub>\u00b0C. Hence, from pressure law.<\/p>\n \u21d2 \\(p_1=p_0\\left(1+\\frac{t_1}{273}\\right)=\\frac{p_0}{273}\\left(273+t_1\\right)=\\frac{p_0}{273} \\times T_1\\)<\/p>\n where \\(T_1=t_1+273\\)<\/p>\n It is dear that the temperature t\u00b0C is the same as the temperature T1<\/sub>K.<\/p>\n Similarly, \\(p_2=\\frac{p_0}{273} \\times T_2\\) [where T2<\/sub> = t2 <\/sub>+ 273]<\/p>\n Here, t2 <\/sub>\u00b0C = T2<\/sub>\u00a0<\/sub>K.<\/p>\n \u2234 \\(\\frac{p_1}{p_2}=\\frac{T_1}{T_2} \\quad \\text { or, } p \\propto T\\), at constant volume.<\/p>\n Hence, pressure law can be expressed as. the pressure of a fixed mass of a gas at constant volume is directly proportional to its temperature in an absolute scale.<\/p>\n Ideal or Perfect gas:<\/strong> The gases that obey Bodes and Charles’ law at any temperature are called ideal gases.<\/p>\n The equation obtained by combining Boyles’s law and Charles’ law is called the equation of the state of an ideal gas.<\/p>\n From Bovle’s law, \\(V \\propto \\frac{1}{p}\\) when T is constant.<\/p>\n From Charles’ law, V\u221dT when p is constant.<\/p>\n \u2234 \\(V \\propto \\frac{T}{p}\\) when both p and T vary<\/p>\n or, pV= kT ….(1)<\/p>\n where k is a constant whose value depends on the units of p, V, T, and the mass of the gas. If a gas of fixed mass occupies a volume V1<\/sub> at pressure p1<\/sub> and temperature T1<\/sub>, and after a general change, a volume V2<\/sub>, at pressure p2<\/sub>, and temperature T2<\/sub>, then<\/p>\n \u21d2 \\(\\frac{p_1 V_1}{T_1}=\\frac{p_2 V_2}{T_2}\\)….(2)<\/p>\n It is known that the physical property of an ideal gas of fixed mass depends on its pressure, volume, and temperature.<\/p>\n Hence, the equation pV = kT is called the equation of state of an ideal gas.<\/p>\n Note that no real gas totally follows this equation of state.<\/p>\n The deviation, though small at ordinary temperature and pressure, becomes significant at high pressure and at low- temperature.<\/p>\n From Charles\u2019 law, If a fixed mass of gas at constant pressure occupies a volume V0<\/sub> at 0\u00b0C, then at t\u00b0C its volume will be \\(V_t=V_0\\left(1+\\frac{t}{273}\\right)\\)<\/p>\n Now if the temperature is reduced to t = -273\u00b0C<\/p>\n \u21d2 \\(V_t=V_0\\left(1-\\frac{273}{273}\\right)=0\\), i.e., the volume of the gas vanishes at -273\u00b0C temperature.<\/p>\n Similarly from pressure law, if a fixed mass of a gas at constant volume has a pressure p at 0\u00b0C, then at t\u00b0C its pressure will be \\(p_t=p_0\\left(1+\\frac{t}{273}\\right) \\text {. }\\)<\/p>\n Now if the temperature is reduced to t = -273 \u00b0C, \\(p_t=p_0\\left(1-\\frac{273}{273}\\right)=0\\), i.e., the pressure of the gas vanishes at -273\u00b0C temperature.<\/p>\n So, at -273 \u00b0C, the volume and pressure of a fixed mass of gas have worked out to be zero.<\/p>\n If the temperature could be lowered below -273 \u00b0C, the pressure and volume would have been negative, which is meaningless. Hence, the lowest conceivable temperature in this universe is -273 \u00b0C.<\/p>\n Therefore -273 \u00b0C is called the absolute zero of temperature, or simply, absolute zero. More precisely, the value of absolute zero is -273.15\u00b0C.<\/p>\n Definition:<\/strong> The temperature at which the volume and the pressure of a gas reduce to zero is called the absolute zero of temperature. This temperature is the lowest temperature in reality.<\/p>\n Taking absolute zero i.e., -273\u00b0C as zero, a new scale of temperature was developed by Lord Kelvin, and it is called the absolute scale of temperature or Kelvin scale after the inventor.<\/p>\n In this scale, the temperature reading is denoted by A (absolute) or K (Kelvin). Each degree in this scale is taken equal to a degree in the Celsius scale. Hence, the scale of temperature where -273\u00b0C is taken as zero (0) and each degree is equal to a degree in Celsius scale, is called the absolute scale or Kelvin scale.<\/p>\n The freezing point of water (0\u00b0C) in an absolute scale is 273 K and the boiling point of water (100\u00b0C) in this scale is 373 K.<\/p>\n If any temperature is represented by t\u00b0C in the Celsius scale and by TK on the Kelvin scale, then, T = t + 273.<\/p>\n In the Kelvin scale, a temperature may be zero or positive. A negative Kelvin temperature does not exist.<\/p>\n 0\u00b0 K In Fahrenheit scale:<\/strong> in the relation \\(\\frac{C}{5}=\\frac{F-32}{9}\\), putting C = -273\u00b0C (0K) we have, \\(\\frac{-273}{5}=\\frac{F-32}{9}\\) or, F = – 459.4\u00b0F.<\/p>\n Incidentally, gas laws are valid as long as the matter remains in a gaseous state. All gases get liquified much before they attain the absolute zero temperature. No gas can be cooled to the temperature of absolute zero, and as such, zero volume or zero pressure of a gas is never realized in practice.<\/p>\n Absolute scale of temperature\u2014why it is so named:<\/strong> The selection of 0 degrees in Celsius and Fahrenheit scale has no scientific reason behind it. The freezing point of water is taken arbitrarily as 0\u00b0C and the boiling point as 100\u00b0C in the Celsius scale.<\/p>\n The equation of state for an ideal gas is pV = kT. The value of the constant k depends on the mass of the gas used. Keeping pressure (p) and temperature (T) constant, if the mass of a gas is doubled, the volume is also doubled.<\/p>\n Hence, for 1 gram-molecule of any ideal gas,<\/p>\n pV = RT ….(1) f the volume of n gram-molecule of gas is V, then the volume of 1 gram-molecule of gas, = V\/n<\/p>\n Then from equation (1) we get, \\(p \\cdot \\frac{V}{n}=R T \\quad \\text { or, } p V=n R T\\)….(2)<\/p>\n Obviously, for mg of a gas of molecular weight M the number of moles, n = m\/n<\/p>\n \u2234 pV = \\(\\frac{m}{M} R T\\)….(3)<\/p>\n Comparing equation(3) with the gas equation pV = kT, k = \\(\\frac{m}{M} R T\\)<\/p>\n Therefore for 1 gram of gas, k = \\(\\frac{R}{M}\\) = r (say), r is also a constant and is called the specific gas constant.<\/p>\n As the molecular weights of different gases are different, the magnitude of this constant is also different for different gases.<\/p>\n Magnitude of universal or molar gas constant:<\/strong> From equation (1) above, the relation R = \\(\\frac{p V}{T}\\) is valid for any ideal gas at any temperature and pressure.<\/p>\n Hence, R is also equal to \\(\\frac{p_0 V_0}{T_0}\\) where p0<\/sub> = 76 cm Hg \u2234 R = \\(\\left(76 \\times 13.596 \\times 980.665 \\mathrm{dyn} \\cdot \\mathrm{cm}^{-2}\\right) \\frac{\\times 22.4 \\times 1000 \\mathrm{~cm}^3 \\cdot \\mathrm{mol}^{-1}}{273 \\mathrm{~K}}\\)<\/p>\n = 8.314 x 107<\/sup> dyn \u00b7 cm \u00b7 mol-1<\/sup>\u00a0\u00b7 K-1<\/sup><\/p>\n = 8.314 x 107<\/sup> erg \u00b7 mol”1 \u00b7 K-1<\/sup><\/p>\n = 8.314 J \u00b7 mol-1<\/sup> \u00b7 K-1<\/sup><\/p>\n Magnitude of specific gas constant:<\/strong> Value of r for agasis r= R\/M.<\/p>\n 1. For hydrogen, molecular weight = 2,<\/p>\n r = \\(frac{R}{2}=\\frac{8.314 \\times 10^7}{2}\\)<\/p>\n = \\(4.16 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{g}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n 2. For oxygen, molecular weight = 32<\/p>\n r = \\(\\frac{R}{32}=\\frac{8.314 \\times 10^7}{32}\\)<\/p>\n = \\(0.26 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{g}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n 3. For nitrogen, molecular weight = 28<\/p>\n r = \\(\\frac{R}{28}=\\frac{8.314 \\times 10^7}{28}\\)<\/p>\n = \\(0.297 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{g}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n 4. For carbon dioxide, molecular weight = 44<\/p>\n r = \\(\\frac{R}{44}=\\frac{8.314 \\times 10^7}{44}\\)<\/p>\n = \\(0.189 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{g}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n Example 1. The mass of 1 litre of hydrogen at STP is 0.0896 g. Calculate the value of R from this data.<\/strong> Volume of 0.0896 g of hydrogen at STP = 1000 cm\u00b3.<\/p>\n Hence, a volume of 2 g or 1 mol of hydrogen at STP = \\(\\frac{1000 \\times 2}{0.0896}=22321.4 \\mathrm{~cm}^3.\\)<\/p>\n Standard pressure = 76x 13.6×981 dyn \u00b7 cm-2<\/sup>; standard temperature = 0\u00b0C = 273 K<\/p>\n \u2234 R = \\(\\frac{p V}{T}=\\frac{76 \\times 13.6 \\times 981 \\times 22321.4}{273} \\mathrm{erg} \\cdot \\mathrm{mol}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n = \\(8.29 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{mol}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n Example 2. The mass of 3.76 litre of oxygen at 2 standard atmo\u00acsphere pressure and 20\u00b0C is 10 g. Find the value of R<\/strong> Given, p = 2 standard atmosphere = 2 x 76 x 13.6 x 981 dyn \u00b7 cm-2<\/sup>,<\/p>\n V = 3.76 x 10\u00b3 cm\u00b3, T = 273 + 20 = 293 K and n = 10\/32 mol, [where 32 is the molecular weight of oxygen]<\/p>\n \u2234 pV=nRT<\/p>\n R = \\(\\frac{p V}{n T}=\\frac{2 \\times 76 \\times 13.6 \\times 981 \\times 3.76 \\times 10^3 \\times 32}{10 \\times 293}\\)<\/p>\n = \\(8.33 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{mol}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n Example 3. The density of air at STP = 1.293 g \u00b7 L-1<\/sup> and that of mercury = 13.6 g \u00b7\u00a0 cm-3<\/sup>. Find the value of the gas constant for 1 g of air.<\/strong> If k is the constant for 1 g of air, then k = pV\/T<\/p>\n According to the problem, 1.293 g of air occupies a volume of 1000 cm\u00b3<\/p>\n \u2234 1g of air occupies a volume of \\(\\frac{1000}{1.293} \\mathrm{~cm}^3\\)<\/p>\n Here, p = \\(76 \\times 13.6 \\times 981 \\mathrm{dyn} \\cdot \\mathrm{cm}^{-1}, T=273 \\mathrm{~K}\\)<\/p>\n \u2234 k = \\(76 \\times 13.6 \\times 981 \\times \\frac{1000}{1.293} \\times \\frac{1}{273}\\)<\/p>\n = \\(0.287 \\times 10^7 \\mathrm{erg} \\cdot \\mathrm{g}^{-1} \\cdot \\mathrm{K}^{-1}\\)<\/p>\n Example 4. The masses, volumes, and pressures of two samples of oxygen and hydrogen gases are equal. Find the ratio of their absolute temperatures.<\/strong> Let the volume of each gas = V, the pressure of each gas = p, and the absolute temperatures of the gases be T1<\/sub> and T2<\/sub> respectively.<\/p>\n If the mass of each sample is m g, then the number of moles of the gases are respectively, \\(n_1=\\frac{m}{32} \\quad \\text { and } n_2=\\frac{m}{2} \\text {. }\\)<\/p>\n As per the equation pV = nRT, for oxygen gas, \\(p V=\\frac{m}{32} R T_1\\) and for hydrogen gas, \\(p V=\\frac{m}{3} R T_2\\)<\/p>\n \u2234 \\(\\frac{m}{32} R T_1=\\frac{m}{2} R T_2 \\quad \\text { or, } \\quad \\frac{T_1}{T_2}=\\frac{32}{2}=16: 1\\)<\/p>\n Boyle’s law:<\/strong> At constant temperature, the volume (V) of a fixed mass of gas is inversely proportional to the pressure (p) of the gas.<\/p>\n i.e \\(V \\propto \\frac{1}{p}\\) or, pV = constant.<\/p>\n Charles’ law:<\/strong> At constant pressure, the volume of a fixed mass of gas increases or decreases by 1\/273 of its volume at 0\u00b0C per degree Celsius rise or fall in temperature.<\/p>\n Pressure law:<\/strong> At constant volume, the pressure of a fixed mass of gas increases or decreases by 1\/273 of its pressure at 0\u00b0C per degree Celsius rise or fall in temperature.<\/p>\n Volume coefficient:<\/strong> The volume coefficient of a fixed mass of a gas at a constant pressure is the increment of its volume per unit volume when its temperature is raised by 1\u00b0C from 0\u00b0C.<\/p>\n Pressure coefficient:<\/strong> The pressure coefficient of a fixed mass of a gas at a constant volume is the increment of its pressure per unit pressure when its temperature is raised by 1\u00b0C from 0\u00b0C.<\/p>\n Absolute zero temperature:<\/strong> The temperature at which the volume and the pressure of an ideal gas theoretically become zero is called the absolute zero temperature. This is the lowest possible temperature in reality.<\/p>\n The scale of temperature in which the temperature -273\u00b0C is taken as zero and each degree interval is equal to a degree interval in the Celsius scale is called the absolute scale of temperature or Kelvin scale.<\/p>\n Absolute zero temperature = 0K = -273\u00b0C = 459.4\u00b0F.<\/p>\n Universal or molar gas constant.<\/p>\n R =8.314 x 107<\/sup> erg \u00b7 mol-1\u00a0<\/sup> K-1<\/sup><\/p>\n = 8.314 J \u22c5 mol-1<\/sup> K-1<\/sup><\/p>\n At constant pressure, the density of a gas is inversely proportional to its absolute temperature.<\/p>\n At constant temperature, the density of a gas is directly proportional to its pressure.<\/p>\n If the volume of a definite mass of a gas is V and its pressure is p, then according to Boyle\u2019s law pV = constant<\/p>\n Let at constant pressure, the volume of some definite mass of a gas at 0\u00b0C be V0<\/sub>. According to Charles\u2019 law the final volume of the gas due to riC rise or fall in temperature,<\/p>\n \u21d2 \\(V_t=V_0\\left(1 \\pm \\frac{t}{273}\\right)\\)<\/p>\n Let at constant volume, the pressure of some definite mass of a gas at 0\u00b0C be p0<\/sub>. According to pressure law. the final pressure of the gas due to t\u00b0C rise or fall in temperature,<\/p>\n \u21d2 \\(p_t=p_0\\left(1 \\pm \\frac{t}{273}\\right)\\)<\/p>\n Volume coefficient of a gas at constant pressure, \\(\\gamma_p=\\frac{V_t-V_0}{V_0 \\cdot t}\\)<\/p>\n Pressure coefficient of a gas at constant volume, \\(\\gamma_v=\\frac{p_t-p_0}{p_0 \\cdot t}\\)<\/p>\n \u21d2 \\(\\gamma_p=\\gamma_v=\\frac{1}{273}{ }^{\\circ} \\mathrm{C}^{-1}\\)<\/p>\n T = t + 273 lf any temperature is represented by t and T in Celsius and Kelvin scales respectively]<\/p>\n For n mol of an ideal gas, pV = nRT = \\(\\frac{m}{M} R T\\) [where m and M are the mass and molecular weight of the gas respectively]<\/p>\n If \u03c1 is the density of an ideal gas, then \\(\\frac{p}{\\rho T}=\\frac{R}{M}\\) [symbols have usual meanings]<\/p>\n Question 1. At what Celsius temperature, does the volume of a gas become zero according to Charles\u2019 law?<\/strong> Question 2. How does the volume of a definite mass of gas change with pressure at constant temperature?<\/strong> Question 3. How does the volume of a definite mass of gas change with its absolute temperature at constant pressure?<\/strong> Question 4. At constant volume, the pressure of a definite mass of gas is directly proportional to its absolute temperature. Is the statement true or false?<\/strong> Question 5. What is the value of the volume coefficient of a gas?<\/strong> Question 6. What is the value of the pressure coefficient of a gas?<\/strong> Direction: These questions have statement 1 and statement 2 Of the four choices given below, choose the one that best describes the two statements.<\/strong><\/p>\n Question 1.<\/strong><\/p>\n Statement 1:<\/strong> Equal masses of helium and oxygen gases are given equal quantities of heat. There will be a greater rise in the temperature of helium compared to that of oxygen.<\/p>\n Statement 2:<\/strong> The molecular weight of oxygen is more than the molecular weight of helium.<\/p>\n Answer:<\/strong> 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.<\/p>\n Question 2.<\/strong><\/p>\n Statement 1:<\/strong> In the upper part of the atmosphere, the temperature of air is of the order of 1000 K, even then it is quite cold there.<\/p>\n Statement 2<\/strong>: Molecular density at high altitudes is low.<\/p>\n Answer:<\/strong>\u00a0 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.<\/p>\n Question 3.<\/strong><\/p>\n Statement 1:<\/strong> Shows the V – T graphs of a certain mass of an ideal gas at two pressures p1<\/sub> and p2<\/sub>. It follows from the graph that p1<\/sub> is greater than p2<\/sub>.<\/p>\n Statement 2:<\/strong> The slope of V- T graph for an ideal gas is directly proportional to pressure.<\/p>\n Answer:<\/strong> 3. Statement 1 is true, statement 2 is false.<\/p>\n Question 4.<\/strong><\/p>\n Statement 1:<\/strong> V-T graph in a process is a rectangular hyperbola. Then p-T graph in the same process will be a parabola.<\/p>\n Statement 2:<\/strong> If the V-T graph is a rectangular hyperbola, with an increase in T, the volume will decrease and hence, pressure will increase.<\/p>\n Answer:<\/strong> 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.<\/p>\n Question 5.<\/strong><\/p>\n Statement 1:<\/strong> The size of a hydrogen balloon increases as it rises in the air.<\/p>\n Statement 2:<\/strong> The material of the balloon can be easily stretched.<\/p>\n Answer:<\/strong> 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.<\/p>\n Question 1. The V-T graph of two gases A and B is shown.<\/strong><\/p>\n Answer:<\/strong> 1. D, 2. D, 3. D, 4. D<\/p>\n Read the following passages carefully and answer the questions at the end of them.<\/strong><\/p>\n Question 1. An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and the temperature at the surface is 40\u00b0C. The atmospheric pressure is 76 cm of Hg and g = 980 cm \u00b7 s-2<\/sup><\/strong>.<\/p>\n 1. What is the pressure at the bottom of the lake?<\/strong><\/p>\n Answer:\u00a0<\/strong> 1. 1279325 dyn \u2022 cm-2<\/sup><\/p>\n 2. What is the temperature at the bottom of the lake?<\/strong><\/p>\n Answer:<\/strong> 2. 10.37\u00b0C<\/p>\n Question 2. A fixed thermally conducting cylinder has a radius of R and a height of L0<\/sub>. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L, from the top surface as shown. The atmospheric pressure is p0<\/sub>.<\/strong><\/p>\n 1. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be<\/strong><\/p>\n Answer:<\/strong> 1. \\(\\frac{p_0}{2}+\\frac{M g}{\\pi R^2}\\)<\/p>\n 2. When the piston is at a distance 2L from the top the hole at the top is sealed. The piston is then released, at a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is<\/strong><\/p>\n\n
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\u00a0Expansion Of Gases Charles Law<\/h2>\n
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<\/p>\nExpansion Of Gases Pressure Law<\/h2>\n
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<\/p>\nExpansion Of Gases – Alternative Forms Of Charles Law And Pressure Law<\/h2>\n
Expansion Of Gases- Combination Of Boyles Law And Charles<\/h2>\n
Expansion Of Gases Absolute Scale Of Temperature<\/h2>\n
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Expansion Of Gases Universal gas Constant<\/h2>\n
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\n(standard pressure) = 76 x 13.596 x 980.665 dyn \u00b7 cm-2<\/sup>, T0<\/sub> (standard temperature) = 0 + 273 = 273 K and V0<\/sub> = the volume at STP of 1 mol of a gas = 22.4 litre, as per Avogadro’s law.<\/p>\nUniversal gas Constant Numerical Examples<\/h2>\n
\nSolution:<\/strong><\/p>\n
\nSolution:<\/strong><\/p>\n
\nSolution:<\/strong><\/p>\n
\nSolution:<\/strong><\/p>\n\u00a0Expansion Of Gases Conclusion<\/h2>\n
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Expansion Of Gases Useful Relations For Solving Numerical Problems<\/h2>\n
Expansion Of Gases Very Short Answer Type Questions<\/h2>\n
\nAnswer:<\/strong> -273\u00b0C<\/p>\n
\nAnswer:<\/strong> Inversely proportional<\/p>\n
\nAnswer:<\/strong> Directly proportional<\/p>\n
\nAnswer:<\/strong> True<\/p>\n
\nAnswer:<\/strong> 1\/273 \u00b0C-1<\/sup><\/p>\n
\nAnswer:<\/strong> 1\/273 \u00b0C-1<\/sup><\/p>\nExpansion Of Gases Assertion Reason Type Question And Answers<\/h2>\n
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![\"Class](\"https:\/\/wbbsesolutions.net\/wp-content\/uploads\/2024\/03\/Class-11-Physics-Unit-7-Properties-Of-Matter-Chapter-4-Thermometry-Graph-Between-Celcius-And-Fahrenhit-Temperature.png\")
<\/p>\nExpansion Of Gases Match Column 1 With Column 2<\/h2>\n
![\"Class](\"https:\/\/wbbsesolutions.net\/wp-content\/uploads\/2023\/11\/Class-11-Physics-Unit-7-Properties-Of-Matter-Chapter-6-Expansion-Of-Gases-V-T-Graph-Of-Two-Gases-A-And-B.png\")
<\/p>\n![\"Class](\"https:\/\/wbbsesolutions.net\/wp-content\/uploads\/2023\/11\/Class-11-Physics-Unit-7-Properties-Of-Matter-Chapter-6-Expansion-Of-Gases-Match-The-Column-Question-1.png\")
<\/p>\nExpansion Of Gases Comprehension Type Questions And Answers<\/h2>\n
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![\"Class](\"https:\/\/wbbsesolutions.net\/wp-content\/uploads\/2023\/11\/Class-11-Physics-Unit-7-Properties-Of-Matter-Chapter-6-Expansion-Of-Gases-A-Fixed-Thermally-Conducting-Cylinder.png\")
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