5.6 L of helium gas at STP is adiabatically | vedclass

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(A) Number of moles $n = \frac{5.6}{22.4} = 0.25 = \frac{1}{4} \ mol$.For helium (monatomic gas),the adiabatic index $\gamma = \frac{5}{3}$.Using the

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$\frac{9}{8}RT_1$

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(A) Number of moles $n = \frac{5.6}{22.4} = 0.25 = \frac{1}{4} \ mol$.For helium (monatomic gas),the adiabatic index $\gamma = \frac{5}{3}$.Using the adiabatic relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$:$T_1 (5.6)^{\frac{5}{3}-1} = T_2 (0.7)^{\frac{5}{3}-1} \Rightarrow T_1 (5.6)^{2/3} = T_2 (0.7)^{2/3}$.$T_2 = T_1 \left( \frac{5.6}{0.7} \right)^{2/3} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4T_1$.The work done by the gas in an adiabatic process is $W_{gas} = \frac{nR(T_1 - T_2)}{\gamma - 1}$.$W_{gas} = \frac{(1/4)R(T_1 - 4T_1)}{(5/3) - 1} = \frac{(1/4)R(-3T_1)}{2/3} = \frac{-3/4 R T_1}{2/3} = -\frac{9}{8}RT_1$.The work done on the gas (external work) is $W_{ext} = -W_{gas} = \frac{9}{8}RT_1$.

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

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