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Standard XII

Chemistry

PV Work

The pressure ...

Question

The pressure and volume of a monoatomic ideal gas are related as $P \propto \frac{1}{V^{2}}$ for a process A $\to$ C as shown in the figure. The pressure and volume at A is $3 P_{0} a n d V_{0}$ respectively and the pressure at C is $P_{0}$ . If the molar heat capacity of the ideal gas for the process $A \to C i n \frac{R}{x}$ Find x. (R gas constant) ___

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Solution

The process is described by
$P = \frac{k}{v^{2}}$ Where k is a constant
Let volume at C be V $\Rightarrow 3 P_{0} V_{0}^{2} = P_{0} V^{2} \Rightarrow V = \sqrt{3} V_{0}$
$W = \int_{V_{0}}^{\sqrt{3} V_{0}} P d V = \int_{V_{0}}^{\sqrt{3} V_{0}} \frac{k}{v^{2}} d V = k {[- \frac{1}{v}]}_{v_{0}}^{\sqrt{3} v_{0}} [\frac{1}{v_{0}} - \frac{1}{\sqrt{3} v_{0}}] = (3 - \sqrt{3}) P_{0} V_{0}$
Change in internal energy $Δ U = \frac{f}{2} n R Δ T = \frac{3}{2} n R Δ T$
$Δ T = T_{c} - T_{A} = \frac{\sqrt{3} P_{0} V_{0}}{n R} - \frac{3 P_{0} V_{0}}{n R} = - (3 - \sqrt{3}) \frac{p_{0} v_{0}}{n R}$
$Δ U = - \frac{3}{2} (3 - \sqrt{3}) P_{0} V_{0}$
First law of thermodynamics, $Δ Q = Δ U + W$
$\Rightarrow Δ Q = - \frac{2}{2} (3 - \sqrt{3}) (P_{0} V_{0} + (3 - \sqrt{3}) P_{0} V_{0} = - \frac{1}{2} (3 - \sqrt{3}) P_{0} V_{0}$
$A n d Δ Q = n C Δ T \Rightarrow n C Δ T = - \frac{1}{2} (3 - \sqrt{3}) P_{0} V_{0}$
$\Rightarrow n . c \frac{(3 - \sqrt{3}) P_{0} V_{0}}{n R} = \frac{(3 - \sqrt{3}) P_{0} V_{0}}{2} \Rightarrow c = \frac{R}{2}$

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