JEE MainPhysicsModerateThermodynamics

P- V diagram of ideal monoatomic gas is as shown. Find net heat
given to the gas in the process ABC. $\$

Show answer and solution

\Delta Q = \Delta U + \Delta W

W_{ABCA} = \frac{1}{2}\left( 2P_{0} \right)\left( 2V_{0} \right) = 2P_{0}V_{0}

W_{ABC} = 2P_{0}V_{0} - W = 2P_{0}V_{0} - \left( - 2P_{0}V_{0} \right) = 4P_{0}V_{0}

\Delta U = \frac{nf}{2}R\Delta T = \frac{f}{2}P\Delta V = \frac{3}{2}P_{0}\left( 2V_{0} \right) = 3P_{0}V_{0}

\therefore\Delta Q = 7P_{0}V_{0}

Source: Grass-G / P-08 Physics paper 23 October.docx

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