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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • 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 \(2V_0\)

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Answer & Solution

Correct Answer

(D) \(\frac{3}{4}\frac{{{P_0}{V_0}}}{R}\)

Step-by-step Solution

Detailed explanation

\(n=1\) mole

\(P=P_{0}\left\{1-\frac{1}{2}\left(\frac{V_{0}}{V}\right)^{2}\right\} \quad ; \quad P V=n R T=R T\)

\(P=\frac{R T}{V}\)

\(\frac{R T}{V}=P_{0}\left\{1-\frac{V_{0}^{2}}{2 V^{2}}\right\}\)

\(T=\frac{P_{0} V}{R}\left\{1-\frac{V^{2}}{2 V^{2}}\right\}=\frac{P_{0}}{R}\left\{V-\frac{V_{0}^{2}}{2 V}\right\}\)

\(\Delta T \)\(=\frac{P_{0}}{R}\left\{\left(2 V_{0}-V_{0}\right)-\frac{V_{0}^{2}}{2}\left(\frac{1}{2 V_{0}}-\frac{1}{V_{0}}\right)\right\}\)

\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}^{2}}{2}\right\} \)

\( \Delta T \)\(=\frac{P_{0}}{R}\left\{\left(2 V_{0}-V_{0}\right)-\frac{V_{0}^{2}}{2}\left(\frac{1}{2 V_{0}}-\frac{1}{V_{0}}\right)\right\} \)

\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}^{2}(1-2)}{2 \times 2 V_{0}}\right\} \)

\(=\frac{P_{0}}{R}\left\{V_{0}-\frac{V_{0}}{4}\right\}=\frac{3}{4} \frac{P_{0} V_{0}}{R}\)

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