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

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(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For helium (a monoatomic gas),th

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

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(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For helium (a monoatomic gas),the adiabatic index $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.Given $V_1 = 5.6 \, 	ext{liter}$ and $V_2 = 0.7 \, 	ext{liter}$,we have $\frac{V_1}{V_2} = \frac{5.6}{0.7} = 8$.Substituting these into the relation: $T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = T_1 (2^2) = 4 T_1$.At $STP$,the number of moles $n = \frac{5.6 \, 	ext{liter}}{22.4 \, 	ext{liter/mol}} = 0.25 = \frac{1}{4} \, 	ext{mol}$.The work done in an adiabatic process is $W = \frac{n R (T_1 - T_2)}{\gamma - 1}$.Substituting the values: $W = \frac{(1/4) R (T_1 - 4 T_1)}{5/3 - 1} = \frac{(1/4) R (-3 T_1)}{2/3} = \frac{-3/4 R T_1}{2/3} = -\frac{9}{8} R T_1$.The magnitude of work done on the gas is $\frac{9}{8} R T_1$.

$5.6 \, \text{liter}$ 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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