Consider one mole of helium gas enclosed in a | vedclass

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(A) For an isothermal process,$P_1 V_1 = P_{	ext{intermediate}} (4 V_1)$,so $P_{	ext{intermediate}} = \frac{P_1}{4}$.For the adiabatic process,$P_{\

Vedclass · Accepted Answer

$1.78$

Vedclass · Answer

(A) For an isothermal process,$P_1 V_1 = P_{	ext{intermediate}} (4 V_1)$,so $P_{	ext{intermediate}} = \frac{P_1}{4}$.For the adiabatic process,$P_{	ext{intermediate}} (4 V_1)^\gamma = P_2 (32 V_1)^\gamma$,where $\gamma = \frac{5}{3}$ for helium.$\frac{P_1}{4} (4 V_1)^{5/3} = P_2 (32 V_1)^{5/3}$$P_2 = \frac{P_1}{4} \left( \frac{4 V_1}{32 V_1} ight)^{5/3} = \frac{P_1}{4} \left( \frac{1}{8} ight)^{5/3} = \frac{P_1}{4} 	imes \frac{1}{32} = \frac{P_1}{128}$.Work done in isothermal process: $W_{	ext{iso}} = nRT \ln \left( \frac{V_f}{V_i} ight) = P_1 V_1 \ln \left( \frac{4 V_1}{V_1} ight) = P_1 V_1 \ln 4 = 2 P_1 V_1 \ln 2$.Work done in adiabatic process: $W_{	ext{adia}} = \frac{P_i V_i - P_f V_f}{\gamma - 1} = \frac{\frac{P_1}{4} (4 V_1) - \frac{P_1}{128} (32 V_1)}{\frac{5}{3} - 1} = \frac{P_1 V_1 - \frac{P_1 V_1}{4}}{\frac{2}{3}} = \frac{\frac{3}{4} P_1 V_1}{\frac{2}{3}} = \frac{9}{8} P_1 V_1$.Ratio: $\frac{W_{	ext{iso}}}{W_{	ext{adia}}} = \frac{2 P_1 V_1 \ln 2}{\frac{9}{8} P_1 V_1} = \frac{16}{9} \ln 2 = f \ln 2$.Therefore,$f = \frac{16}{9} \approx 1.7778 \approx 1.78$.

List-$I$	List-$II$
$A$. Isothermal Process	$I$. Work done by the gas decreases internal energy
$B$. Adiabatic Process	$II$. No change in internal energy
$C$. Isochoric Process	$III$. The heat absorbed goes partly to increase internal energy and partly to do work
$D$. Isobaric Process	$IV$. No work is done on or by the gas

Similar Questions

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