One mole of an ideal gas left( frac{C_P}{C_V} | vedclass

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(D) The molar heat capacity $C$ is given by $C = C_V + \frac{dW}{n dT}$.For an ideal gas,$C_V = \frac{R}{\gamma - 1}$.The work done in the process $P

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$C = \frac{R(\gamma + 1)}{2(\gamma - 1)}$

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(D) The molar heat capacity $C$ is given by $C = C_V + \frac{dW}{n dT}$.For an ideal gas,$C_V = \frac{R}{\gamma - 1}$.The work done in the process $P = \alpha V$ is $W = \int P dV = \int \alpha V dV = \frac{1}{2} \alpha (V_f^2 - V_i^2) = \frac{1}{2} (P_f V_f - P_i V_i)$.Using the ideal gas law $PV = nRT$,we have $W = \frac{1}{2} nR (T_f - T_i) = \frac{1}{2} nR \Delta T$.Thus,the work done per mole per unit temperature change is $\frac{W}{n \Delta T} = \frac{R}{2}$.Therefore,$C = \frac{R}{\gamma - 1} + \frac{R}{2} = R \left( \frac{1}{\gamma - 1} + \frac{1}{2} \right) = \frac{R(\gamma + 1)}{2(\gamma - 1)}$.

One mole of an ideal gas $\left( \frac{C_P}{C_V} = \gamma \right)$ is heated according to the law $P = \alpha V$,where $P$ is the pressure of the gas,$V$ is the volume,and $\alpha$ is a constant. What is the molar heat capacity of the gas in this process?

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