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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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