5.6 L of helium gas at STP is adiabatically | vedclass

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(A) For an adiabatic process,the work done is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Helium is a monatomic gas,so $\gamma = \frac{5}{3}$.The

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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 $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Helium is a monatomic gas,so $\gamma = \frac{5}{3}$.The number of moles $n = \frac{V_1}{22.4} = \frac{5.6}{22.4} = \frac{1}{4} \ mol$.The adiabatic relation is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_1 (5.6)^{\frac{5}{3}-1} = T_2 (0.7)^{\frac{5}{3}-1} \implies T_1 (5.6)^{2/3} = T_2 (0.7)^{2/3}$.$T_2 = T_1 \left( \frac{5.6}{0.7} \right)^{2/3} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4 T_1$.Now,$W = \frac{(1/4) R (T_1 - 4 T_1)}{(5/3) - 1} = \frac{(1/4) R (-3 T_1)}{2/3} = \frac{-3/4 R T_1}{2/3} = -\frac{9}{8} R T_1$.The magnitude of work done on the gas is $\frac{9}{8} R T_1$.

$5.6 \ L$ 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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