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(C) Given $n = 1$ mole.From the ideal gas equation,$PV = nRT = RT$,so $P = \frac{RT}{V}$.Substituting this into the given relation $P = P_0 \left[ 1 -

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$\frac{5}{4} \frac{P_0 V_0}{R}$

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(C) Given $n = 1$ mole.From the ideal gas equation,$PV = nRT = RT$,so $P = \frac{RT}{V}$.Substituting this into the given relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$:$\frac{RT}{V} = P_0 \left[ 1 - \frac{V_0^2}{2V^2} \right]$$T = \frac{P_0}{R} \left[ V - \frac{V_0^2}{2V} \right]$Now,calculate temperature at $V = V_0$ and $V = 2V_0$:$T_1 = T(V_0) = \frac{P_0}{R} \left[ V_0 - \frac{V_0^2}{2V_0} \right] = \frac{P_0}{R} \left[ V_0 - \frac{V_0}{2} \right] = \frac{P_0 V_0}{2R}$$T_2 = T(2V_0) = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0^2}{2(2V_0)} \right] = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0}{4} \right] = \frac{P_0}{R} \left[ \frac{7V_0}{4} \right] = \frac{7 P_0 V_0}{4R}$Change in temperature $\Delta T = T_2 - T_1 = \frac{7 P_0 V_0}{4R} - \frac{P_0 V_0}{2R} = \frac{7 P_0 V_0 - 2 P_0 V_0}{4R} = \frac{5 P_0 V_0}{4R}$.

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$. Here $P_0$ and $V_0$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_0$ to $2V_0$.

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