Two moles of helium gas left(gamma = frac{5}{ | vedclass

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(A) Initial state: $n = 2 \ mol$,$T_1 = 27^{\circ} C = 300 \ K$,$P_1 = P$,$V_1 = V$. Step $1$ (Isobaric expansion): Volume doubles,so $V_2 = 2V$. Sinc

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$7470$

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(A) Initial state: $n = 2 \ mol$,$T_1 = 27^{\circ} C = 300 \ K$,$P_1 = P$,$V_1 = V$. Step $1$ (Isobaric expansion): Volume doubles,so $V_2 = 2V$. Since $P$ is constant,$\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies T_2 = T_1 \left(\frac{V_2}{V_1}ight) = 300 	imes 2 = 600 \ K$. Step $2$ (Adiabatic process): The gas expands/compresses until $T_3 = T_1 = 300 \ K$. Work done in an adiabatic process is $W = \frac{nR(T_2 - T_3)}{\gamma - 1}$. Given $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$. $W = \frac{2 	imes 8.3 	imes (600 - 300)}{2/3} = \frac{2 	imes 8.3 	imes 300 	imes 3}{2} = 8.3 	imes 900 = 7470 \ J$.

Column-$1$	Column-$2$
$P$: Process-$I$	$A$: Adiabatic
$Q$: Process-$II$	$B$: Isobaric
$R$: Process-$III$	$C$: Isochoric
$S$: Process-$IV$	$D$: Isothermal

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