A given mass of a gas is compressed isotherma | vedclass

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(B) In an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P \propto \frac{1}{V}$.Let the initial pressur

Vedclass · Accepted Answer

$1.41$

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(B) In an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P \propto \frac{1}{V}$.Let the initial pressure be $P$ and initial volume be $V_1$. After isothermal compression,pressure $P_2 = 2P$ and volume $V_2 = V_1/2$.Thus,$\frac{V_1}{V_2} = 2$.Now,the gas expands adiabatically to its original volume $V_1$. Let the final pressure be $P_3 = 0.75P$.For an adiabatic process,$P_2 V_2^\gamma = P_3 V_1^\gamma$.Substituting the values: $(2P) \left(\frac{V_1}{2}\right)^\gamma = (0.75P) V_1^\gamma$.$2 \cdot \left(\frac{1}{2}\right)^\gamma = 0.75$.$2^{1-\gamma} = \frac{3}{4} = 3 \cdot 2^{-2}$.$2^{3-\gamma} = 3$.Taking log on both sides: $(3-\gamma) \log 2 = \log 3$.$3-\gamma = \frac{\log 3}{\log 2} \approx 1.585$.$\gamma = 3 - 1.585 = 1.415 \approx 1.41$.

$A$ given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be $0.75$ of its initial pressure. The ratio of the specific heats of the gas is approximately:

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