- JEE Mains 2019
- Hard
- MCQ
One mole of an ideal gas passes through a process where pressure and volume obey the relation \(P\, = {P_0}\,\left[ {1 - \frac{1}{2}{{\left( {\frac{{{V_0}}}{V}} \right)}^2}} \right]\). Here \(P_0\) and \(V_0\) are constants. Calculate the change in the temperature of the gas if its volume change from \(V_0\) to \(2V_0\)
Answer & Solution
(D) \(\frac{3}{4}\frac{{{P_0}{V_0}}}{R}\)
Step-by-step Solution
Detailed explanation
\(P=P_{0}\left\{1-\frac{1}{2}\left(\frac{V_{0}}{V}\right)^{2}\right\} \quad ; \quad P V=n R T=R T\)
\(P=\frac{R T}{V}\)
\(\frac{R T}{V}=P_{0}\left\{1-\frac{V_{0}^{2}}{2 V^{2}}\right\}\)
\(T=\frac{P_{0} V}{R}\left\{1-\frac{V^{2}}{2 V^{2}}\right\}=\frac{P_{0}}{R}\left\{V-\frac{V_{0}^{2}}{2 V}\right\}\)
\(\Delta T \)\(=\frac{P_{0}}{R}\left\{\left(2 V_{0}-V_{0}\right)-\frac{V_{0}^{2}}{2}\left(\frac{1}{2 V_{0}}-\frac{1}{V_{0}}\right)\right\}\)
\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}^{2}}{2}\right\} \)
\( \Delta T \)\(=\frac{P_{0}}{R}\left\{\left(2 V_{0}-V_{0}\right)-\frac{V_{0}^{2}}{2}\left(\frac{1}{2 V_{0}}-\frac{1}{V_{0}}\right)\right\} \)
\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}^{2}(1-2)}{2 \times 2 V_{0}}\right\} \)
\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}}{4}\right\}=\frac{3}{4} \frac{P_{0} V_{0}}{R}\)
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