A given mass of a gas is compressed isotherma | vedclass

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(B) For an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P_1 V_1 = P_2 V_2$.Given that the pressure is

Vedclass · Accepted Answer

$1.41$

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(B) For an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P_1 V_1 = P_2 V_2$.Given that the pressure is doubled,$P_2 = 2 P_1$.Thus,$P_1 V_1 = (2 P_1) V_2$,which gives $\frac{V_1}{V_2} = 2$.Now,the gas expands adiabatically until its original volume is restored,so $V_3 = V_1$.The adiabatic process follows $P V^\gamma = 	ext{constant}$,so $P_2 V_2^\gamma = P_3 V_3^\gamma$.Given $P_3 = 0.75 P_1$ and $P_2 = 2 P_1$,we have:$(2 P_1) V_2^\gamma = (0.75 P_1) V_1^\gamma$.Dividing both sides by $P_1$,we get $2 V_2^\gamma = 0.75 V_1^\gamma$.$\left(\frac{V_1}{V_2}ight)^\gamma = \frac{2}{0.75} = \frac{2}{3/4} = \frac{8}{3}$.Since $\frac{V_1}{V_2} = 2$,we have $2^\gamma = \frac{8}{3} = 2.667$.Taking the logarithm on both sides: $\gamma \log 2 = \log 2.667$.$\gamma = \frac{\log 2.667}{\log 2} \approx \frac{0.426}{0.301} \approx 1.41$.

$A$ 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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