The figure shows a process AB undergone by 2 | vedclass

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(C) Given the process $VT = 	ext{constant}$.Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.Substituting this into the process equatio

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The work done by the gas is $-400R 	ext{ J}$.

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(C) Given the process $VT = 	ext{constant}$.Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.Substituting this into the process equation: $V \left( \frac{PV}{nR} ight) = 	ext{constant} \implies PV^2 = 	ext{constant}$.This is a polytropic process $PV^x = 	ext{constant}$ with $x = 2$.For a diatomic gas,the molar heat capacity at constant volume is $C_V = \frac{5R}{2}$.The molar heat capacity for a polytropic process is $C = C_V + \frac{R}{1-x}$.$C = \frac{5R}{2} + \frac{R}{1-2} = \frac{5R}{2} - R = \frac{3R}{2} 	ext{ J/mol-K}$.The work done in a polytropic process is $W = \frac{nR(T_2 - T_1)}{1-x}$.Given $n = 2$,$T_1 = 300 	ext{ K}$,$T_2 = 500 	ext{ K}$,and $x = 2$:$W = \frac{2R(500 - 300)}{1-2} = \frac{2R(200)}{-1} = -400R 	ext{ J}$.Thus,option $C$ is correct.

The figure shows a process $AB$ undergone by $2$ moles of an ideal diatomic gas. The process $AB$ follows the relation $VT = \text{constant}$. Given $T_1 = 300 \text{ K}$ and $T_2 = 500 \text{ K}$ ($R$ is the gas constant).

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