## Expansion Of Solid And Liquids Introduction Expansion Of Liquids

Liquids expand with the rise in temperature just like solids. The solids have definite shapes. Hence, in case of solids three types of expansion (linear, surface and volume) are significant.

But liquids have no definite shape. Thus unlike solids, liquid expansion with change in temperature is studied only in terms of the change in volume.

- Other characteristic features in liquid expansion are CD For the same rise in temperature, thermal expansion of a liquid is about ten times that of a solid of the same volume.
- For the same rise in temperature, different liquids of equal volume expand differently.
- The rate of thermal expansion differs a little for the different ranges of temperature change.

**Example:** Expansion of water is different for temperature ranges 10°C to 11°C and 93°C to 94°C. Moreover, water contracts in volume when temperature increases from 0°C to 4°C.

**Read and Learn More: Class 11 Physics Notes**

## Expansion Of Solid And Liquids Apparent And Real Expansion Of Liquids Numerical Examples

**Example 1. The coefficient of apparent expansion of mercury with respect to glass is 153 x 10 ^{-6} °C^{-1}. Find the coefficient of linear expansion (α_{g}) of glass where coefficient of real expansion of mercury is 180 x 10^{-6} °C^{-1}.**

**Solution:**

Since coefficient of real expansion (γ) of mercury = coefficient of apparent expansion (γ’) of mercury + coefficient of volume expansion of the material of the container (γ_{g}),

180 x 10^{-6} = 153 x 10^{-6} + γ_{g}

or, γ_{g} = (180- 153) x 10^{-6} = 27 x 10^{-6} °C^{-1}

Also, γ_{g} = 3α_{g} [α_{g} = coefficient of linear expansion of the material of the container]

∴ \(\alpha_g=\frac{\gamma_g}{3}=\frac{27}{3} \times 10^{-6}=9 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

**Example 2. A liquid has a coefficient of apparent expansion, 18 x 10 ^{-5} °C^{-1} for an iron container and 14.46 x 10^{-5} °C^{-1} for an aluminium container. If the coefficient of linear expansion of aluminium is 2.38 x 10^{-5} °C^{-1}, find that of iron.**

**Solution:**

We know γ =γ’ + γ_{g}, where, γ = coefficient of real expansion of the liquid, γ’ = coefficient of apparent expansion of the liquid,γ_{g} = coefficient of volume expression of the material of the container.

⇒ \(\gamma=18 \times 10^{-5}+\gamma_{\text {iron }}\)….(1)

and for aluminium container, \(\gamma=14.46 \times 10^{-5}+\gamma_{\mathrm{Al}}\)

= \(14.46 \times 10^{-5}+3 \times 2.38 \times 10^{-5 \circ} \mathrm{C}^{-1}\)

⇒ \({\left[because \gamma_{\mathrm{Al}}=3 \alpha_{\mathrm{Al}}\right]}\)

= \(21.60 \times 10^{-5 \circ} \mathrm{C}^{-1}\)…(2)

From (1) and (2), we get,

⇒ \(\gamma_{\text {iron }}=21.60 \times 10^{-5}-18 \times 10^{-5}=3.6 \times 10^{-5}\)

or, \(3 \alpha_{\text {iron }}=3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)

∴ \(a_{\text {iron }}=1.2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)

**Example 3. A thin long glass tube of uniform cross-section contains 1 m long mercury thread at 0°C. At 100°C, the mercury thread increases by 16.5 mm. If the coefficient of real expansion of mercury is 0. 000182 °C ^{-1}, find the coefficient of linear expansion of glass.**

**Solution:**

Let the areas of cross-section of the glass tube be A_{0} cm² and A_{100 }cm² at 0°C and 100°C respectively.

Volume of mercury at 0°C, V_{0} = 100 x A_{0} cm³…..(1)

and volume of mercury at 100°C, \(V_{100}=101.65 \times A_{100} \mathrm{~cm}^3\)….(2)

Also from surface expansion of glass we get,

⇒ \(A_{100}=A_0(1+100 \beta)\)

= \(A_0\left(1+100 \times 2 \alpha_g\right)\)…(3)

From definition, γ for mercury = \(\frac{V_{100}-V_0}{V_0 \times 100}\)….(4)

Substituting the values of γ, V_{0}, V_{100} and using (3), we have, 1

0.000182 = \(\frac{101.65 \times A_{100}-100 \times A_0}{100 \times A_0 \times 100}\)

= \(\frac{101.65 A_0\left(1+200 \alpha_g\right)-100 A_0}{100 \times A_0 \times 100}\)

= \(\frac{101.65\left(1+200 \alpha_g\right)-100}{10^4}\)

or, \(1.82=101.65(1+200 \alpha)-100\)

∴ \(1+200 \alpha=\frac{101.82}{101.65}\)

or, \(\alpha=8.3 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} .\)

**Example 4. The internal volume of a glass flask is V cm³. What** **volume of mercury should be kept in the flask so that the volume of the empty space over mercury in the flask remains constant at all temperatures? Coefficient of volume expansion of mercury = 1.8 x 10 ^{-4} °C^{-1} and coefficient of linear expression of glass = 9 x 10^{-6} °C^{-1}.**

**Solution:**

Let the required volume of mercury be x cm³.

To keep the volume of empty space constant at all temperatures, the expansion of mercury should be equal to that of glass for the same rise in temperature.

If the temperature rise is t°C, \(x \times 1.8 \times 10^{-4} \times t=V \times 3 \times 9 \times 10^{-6} \times t\)

or, x = \(\frac{V \times 27 \times 10^{-6}}{1.8 \times 10^{-4}}=\frac{27}{180} V=\frac{3}{20} V\)

∴ 3/20 th part of the flask should be filled up with mercury.

**Example 5. The volume expansion coefficients of glass and mercury are 2.4 x 10 ^{-5 }°C^{-1} and 1.8 x 10^{-4} °C^{-1} respectively. What volume of mercury should be kept in the flask so that the volume of the empty space over mercury in the flask remains constant at all temperatures?**

**Solution:**

Let the required volume of mercury = x cm³ and volume of the glass container = V cm³.

To keep the volume of empty space constant at any temperature, the expansion of mercury should be equal to that of the glass container for the same rise in temperature.

If the temperature rise is t°C, \(x \times 1.8 \times 10^{-4} \times t=V \times 2.4 \times 10^{-5} \times t\)

or, \(x=\frac{V \times 2.4 \times 10^{-5}}{1.8 \times 10^{-4}}=V \times \frac{24}{180}=\frac{2}{15} V\)

∴ 2/15 th part of the flask should be filled with mercury.

**Example 6. The internal volume of a glass flask is 540 cm³. What volume of mercury should be kept in the flask so that the volume of the empty space remains constant at any temperature? Real expansion of mercury = 1.8 x 10 ^{-4 °}C^{-1} and volume expansion of glass = 2.5 x 10^{-5} °C^{-1}.**

**Solution:**

Let the volume of mercury be x cm³.

To keep the volume of the empty space constant at any temperature, the expansion of mercury should be equal to that of the glass flask for the same rise in temperature. I the temperature rise is t°C than,

x x 1.8 x 10^{-4 }x t = 540 x 2.5 x 10^{-5} x t or, x = 75

∴ 75 cm³ of mercury should be kept in the flask

**Example 7. The volume of the bulb of a mercury thermometer is 1 cm³ at 0°C, and it is filled with mercury at that temperature. The tube attached to the bulb has an area of cross-section 0.1 mm². If the coefficient of apparent expansion of mercury is 16 x 10 ^{-5} °C^{-1}, find the length up to which mercury expands in the tube, when the bulb is immersed in boiling water.**

**Solution:**

The expansion of mercury in the bulb = initial vol¬ume of mercury x its coefficient of apparent expansion x increase in temperature

= 1 x 16 x 10^{-5} x (100-0) = 16 x 10^{-3} cm³

Suppose the mercury thread expands by a length x in the tube.

∴ x x 0.1 x 10^{-2} = 16 x 10^{-3}

∴ x = 16 cm.

**Example 8. 1 g water has a volume of 1 cm³ at 4°C. Its volume at 60°C is 1.0169 cm³. Calculate the average coefficient of real expansion of water between these two temperatures.**

**Solution:**

Coefficient of real expansion \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)

= \(\frac{1.0169-1}{1 \times(60-4)}=\frac{0.0169}{56}=3.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1} .\)

**Example 9. Masses of 10 cm³ of water are 9.998 g and 10 g at 0°C and 4°C respectively. Find the average coefficient of real expansion of water within the range of these temperatures.**

**Solution:**

Density of water at 0°C, \(\rho_0=\frac{9.998}{10}=0.9998 \mathrm{~g} \cdot \mathrm{cm}^{-3} .\)

Density of water at 4°C, \(\rho_4=\frac{10}{10}=1 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Therefore, the average coefficient of real expansion of water within the range of 0°C to 4°C,

⇒ \(\gamma=\frac{\rho_0-\rho_4^{100}}{\rho_0 \times t} \frac{0.9998-1}{0.9998 \times(4-0)}=-5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)

The negative value of γ indicates that water contracts when it is heated from 0°C to 4°C.

**Example 10. Density of mercury at 0°C is 13.5955 g · cm ^{-3}. If the coefficient of linear expansion of mercury is 0. 000061 °C^{-1}, find its density at 60°C.**

**Solution:**

Given, \(\rho_0=13.5955 \mathrm{~g} \cdot \mathrm{cm}^{-3}, \rho_{60}=\)?

⇒ \(\gamma=3 \alpha=3 \times 61 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} \text {. }\)

Density of mercury at \(60^{\circ} \mathrm{C}\),

⇒ \(\rho_{60}=\frac{\rho_0}{1+\gamma t}=\frac{13.5955}{1+3 \times 61 \times 10^{-6} \times 60}\)

⇒ \({\left[because t=60^{\circ}-0^{\circ}=60^{\circ}\right]}\)

=13.448 g · cm³

**Example 11. Density of mercury at 15°C is 13.56 g · cm ^{-3} and its coefficient of real expansion is 18 x 10^{-5 °}C^{-1}. What will be the mass of 600 cm³ of mercury at 130°C? What will be the volume of 600 g of mercury at 130°C?**

**Solution:**

We know, \(\rho_2=\rho_1\left[1-\gamma\left(t_2-t_1\right)\right]\)

Here density of mercury at 15°C, ρ_{15} = 13.56 g · cm^{-3}

Coefficient of real expansion, γ= 18 x 10^{-5} °C^{-1}; t_{1} = 15°C and t_{2} = 130°C

Increase in temperature, t = 130-15 = 115°C

∴ Density of mercury at 130 °C,

⇒ \(\rho_{130}=\frac{\rho_{15}}{1+\gamma t}=\frac{13.56}{1+18 \times 115 \times 10^{-5}}=13.285 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

∴ Mass of 600 cm³ of mercury at \(130^{\circ} \mathrm{C}\)

= \(600 \times 13.285=7971 \mathrm{~g}\)

Volume of \(600 \mathrm{~g}\) of mercury at \(130^{\circ} \mathrm{C}\)

= \(\frac{600}{13.285}=45.163 \mathrm{~cm}^3 \text {. }\)

**Example 12. Density of mercury is 13.6 g · cm ^{-3} at 0°C. What will be the volume of 100 g of mercury at 100°C? Coefficient of real expansion of mercury 1/5550 °C^{-1}.**

**Solution:**

Here, density of mercury at 0°C, ρ_{0} = 13.6 g · cm^{-3}

Coefficient of real expansion, \(\gamma=\frac{1}{5550}{ }^{\circ} \mathrm{C}^{-1}\)

Increase in temperature, t = 100-0 = 100°C

∴ Density of mercury at 100°C, \(\rho_{100}=\frac{\rho_0}{1+\gamma t}=\frac{13.6}{1+\frac{100}{5550}}=\frac{13.6 \times 5550}{5650}\)

∴ Volume of 100 g of mercury at 100°C = \(\frac{100}{13.359}=7.485 \mathrm{~cm}^3\)