Ncert Exercises & Solutions

Consider a p-V diagram in which the path followed by one mole of a perfect gas in a cylindrical container is shown in the figure.

(a) Find the work done when the gas is taken from state 1 to state 2.

(b) What is the ratio of temperatures $T_{1} / T_{2}$ if $V_{2} = 2 V_{1}$ ?
(c) Given the internal energy for one mole of gas at temperature T is (3/2)RT, find the heat supplied to the gas when it is taken from state 1 to 2 with $V_{2} = 2 V_{1}$ .

Hint: The process is an adiabatic process.

Let,

p V^{1 / 2}

=Constant=K, p=

\frac{K}{\sqrt{V}}

Step 1: Find the work done by the gas.

(a) Work done for the process 1 to 2,

W = \int_{V_{1}}^{V_{2}} p d V = K \int_{V_{1}}^{V_{2}} \frac{d V}{\sqrt{V}} = K {[\frac{\sqrt{V}}{1 / 2}]}_{V_{1}}^{V_{2}} = 2 K (\sqrt{V_{2}} - \sqrt{V_{1}})

= 2 p_{1} {V_{1}}^{1 / 2} (\sqrt{V_{2}} - \sqrt{V_{1}}) = 2 p_{2} {V_{2}}^{1 / 2} (\sqrt{V_{2}} - \sqrt{V_{1}})

Step 2: Use the ideal gas equation.

(b) From the ideal gas equation,

p V = n R T \Rightarrow T = \frac{p V}{n R} = \frac{p \sqrt{V} \sqrt{V}}{n R}

\Rightarrow T = \frac{K \sqrt{V}}{n R} (A s, p \sqrt{V} = K)

Hence,

T_{1} = \frac{K \sqrt{V_{1}}}{n R} & T_{2} = \frac{K \sqrt{V_{2}}}{n R}

\Rightarrow \frac{T_{1}}{T_{2}} = \frac{\frac{K \sqrt{V_{1}}}{n R}}{\frac{K \sqrt{V_{2}}}{n R}} = \sqrt{\frac{V_{1}}{V_{2}}} = \sqrt{\frac{V_{1}}{2 V_{1}}} = \frac{1}{\sqrt{2}} (∵ V_{2} = 2 V_{1})

Given, the internal energy of the gas,

U = (\frac{3}{2}) R T

∆ U = U_{2} - U_{1} = \frac{3}{2} R (T_{2} - T_{1})

= \frac{3}{2} R T_{1} (\sqrt{2} - 1)

[∵ T_{2} = \sqrt{2} T_{1}]

∆ W = 2 p_{1} {V_{1}}^{1 / 2} (\sqrt{V_{2}} - \sqrt{V_{1}})

= 2 p_{1} {V_{1}}^{1 / 2} (\sqrt{2} \times \sqrt{V_{1}} - \sqrt{V_{1}})

= 2 p_{1} V_{1} (\sqrt{2} - 1) = 2 R T_{1} (\sqrt{2} - 1)

∵ ∆ Q = ∆ U + ∆ W = \frac{3}{2} R T_{1} (\sqrt{2} - 1) + 2 R T_{1} (\sqrt{2} - 1)

= (\sqrt{2} - 1) R T_{1} (2 + 3 / 2)

= (\frac{7}{2}) R T_{1} (\sqrt{2} - 1)

This is the amount of heat supplied.

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