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(B) For an adiabatic process,the work done $W$ is given by the first law of thermodynamics: $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$

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$W = \mu C_V(T_1 - T_2)$

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(B) For an adiabatic process,the work done $W$ is given by the first law of thermodynamics: $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$,so $W = -\Delta U$.The change in internal energy is $\Delta U = \mu C_V(T_2 - T_1)$.Therefore,$W = -\mu C_V(T_2 - T_1) = \mu C_V(T_1 - T_2)$.Alternatively,using the adiabatic work formula: $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.Since $R = C_P - C_V$ and $\gamma = \frac{C_P}{C_V}$,we have $\gamma - 1 = \frac{C_P - C_V}{C_V} = \frac{R}{C_V}$.Substituting this into the work formula: $W = \frac{\mu R(T_1 - T_2)}{R / C_V} = \mu C_V(T_1 - T_2)$.

An ideal gas undergoes an adiabatic process from state $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$. The work done during the process is ..... (where $\mu$ = number of moles,$C_P$ and $C_V$ = molar specific heats).

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