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(D) For an isobaric process,the heat supplied is given by $\Delta Q = nC_p \Delta T$. According to the first law of thermodynamics,$\Delta Q = \Delta

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$\frac{\gamma-1}{\gamma}$

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(D) For an isobaric process,the heat supplied is given by $\Delta Q = nC_p \Delta T$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $\Delta U = nC_v \Delta T$ is the change in internal energy and $W$ is the work done. Thus,$W = \Delta Q - \Delta U = nC_p \Delta T - nC_v \Delta T = n(C_p - C_v) \Delta T$. Taking the ratio $\frac{W}{\Delta Q}$,we get: $\frac{W}{\Delta Q} = \frac{n(C_p - C_v) \Delta T}{nC_p \Delta T} = \frac{C_p - C_v}{C_p} = 1 - \frac{C_v}{C_p}$. Since $\gamma = \frac{C_p}{C_v}$,we have $\frac{C_v}{C_p} = \frac{1}{\gamma}$. Therefore,$\frac{W}{\Delta Q} = 1 - \frac{1}{\gamma} = \frac{\gamma - 1}{\gamma}$.

In an isobaric process of an ideal gas,the ratio of work done by the system to the heat supplied $\left(\frac{W}{Q}\right)$ is

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