Question

# Two samples A and B of same gas have same initial pressure and volume. A undergoes an Isothermal expansion and B is subjected to adiabatic expansion. If the final volume in both cases is double the initial volume and work done in both the cases are same, then prove that$$2^{1 - \gamma} - 1 = ln2, \gamma = \dfrac{C_p}{C_\gamma}$$

Solution

## Let the initial pressure, $$p_{i} = p_{1}$$Initial volume, $$V_{i} = V_{1}$$final volume, $$V_{f} = V_{2}$$$$= 2V_{1}$$Work done in isothermal process,$$W_{iso} = 2.303 \,RT \,log \left ( \dfrac{V_2}{V_1} \right )$$Since $$PV = RT$$ (1 mol of gas),$$W_{iso} = P_1V_1ln_2$$ ..........(1)work done in adiabatic process,$$W_{ad} = \dfrac{1}{\gamma - 1} [P_1V_1 - P_2V_2]$$$$= \dfrac{P_1V_1}{\gamma - 1} \left [ 1 - \dfrac{P_2V_2}{P_1V_1} \right ]$$For adiabatic process,$$P_{1}V_{1}^{\gamma} = P_{1}V_{2}^{\gamma} = K (constant)$$$$\therefore \dfrac{P_2}{P_1} = \dfrac{V_1^\gamma}{V_2^\gamma}$$$$\therefore W_{ad} = \dfrac{P_1V_1}{\gamma - 1} \left [ 1 - \dfrac{V_2^{1 - \gamma}}{V_1^{1 - \gamma}} \right ]$$$$W_{ad} = \dfrac{P_1V_1}{\gamma - 1} [1 - 2^{1 - \gamma}]$$ .............(2)since work done is same for both processes,$$W_{iso} = W_{ad}$$$$\therefore P_{1}V_{1}l 2 = \dfrac{P_1V_1}{\gamma - 1} (1 - 2^{1 - \gamma})$$ .........(3)Re-arranging and simplifying (3),$$2^{1 - \gamma} - 1 = (1 - \gamma) ln2.$$Physics

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