5.6 L of Helium gas at STP is compressed adi | vedclass

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(A) For Helium (monatomic gas),the adiabatic index $\gamma = 5/3$. At $STP$,the volume of $1 \ mole$ of gas is $22.4 \ L$. Number of moles $n = \frac{

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$\frac{9}{8} RT_1$

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(A) For Helium (monatomic gas),the adiabatic index $\gamma = 5/3$. At $STP$,the volume of $1 \ mole$ of gas is $22.4 \ L$. Number of moles $n = \frac{5.6}{22.4} = \frac{1}{4} \ mole$. For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$. Given $V_1 = 5.6 \ L$ and $V_2 = 0.7 \ L$,so $V_1/V_2 = 8$. $T_2 = T_1 (V_1/V_2)^{\gamma-1} = T_1 (8)^{5/3 - 1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4T_1$. Work done $W = -\frac{nR(T_2 - T_1)}{\gamma - 1}$. Substituting the values: $W = -\frac{(1/4)R(4T_1 - T_1)}{5/3 - 1} = -\frac{(1/4)R(3T_1)}{2/3} = -\frac{3}{4} R T_1 	imes \frac{3}{2} = -\frac{9}{8} RT_1$. The magnitude of work done is $\frac{9}{8} RT_1$.

$5.6 \ L$ of Helium gas at $STP$ is compressed adiabatically to $0.7 \ L$. If the initial temperature is $T_1$,then the work done during the process is...?

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