5.6 text{ liter} of helium gas at STP is adia | vedclass

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Question

(A) For an adiabatic process,the work done is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Given: Initial volume $V_1 = 5.6 	ext{ L}$,final volume

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$\frac{9}{8} R T_1$

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(A) For an adiabatic process,the work done is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Given: Initial volume $V_1 = 5.6 	ext{ L}$,final volume $V_2 = 0.7 	ext{ L}$.Helium is a monoatomic gas,so $\gamma = \frac{5}{3}$.The number of moles $n = \frac{5.6 	ext{ L}}{22.4 	ext{ L}} = \frac{1}{4} 	ext{ mol}$.Using the adiabatic relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$:$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = T_1 \left( \frac{5.6}{0.7} ight)^{\frac{5}{3}-1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4T_1$.Now,calculate the work done $W$ (work done on the gas is positive in this convention,or $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$):$W = \frac{\frac{1}{4} R (T_1 - 4T_1)}{\frac{5}{3} - 1} = \frac{\frac{1}{4} R (-3T_1)}{\frac{2}{3}} = \frac{1}{4} R (-3T_1) 	imes \frac{3}{2} = -\frac{9}{8} R T_1$.The magnitude of work done is $\frac{9}{8} R T_1$.

$5.6 \text{ liter}$ of helium gas at $STP$ is adiabatically compressed to $0.7 \text{ liter}$. Taking the initial temperature to be $T_1$,the work done in the process is

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