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(A) For an adiabatic process, the heat exchange $\Delta Q = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. Since

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$25$

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(A) For an adiabatic process, the heat exchange $\Delta Q = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. Since $\Delta Q = 0$, we have $\Delta U = -\Delta W$. Given work done by the gas $\Delta W = 83 \,J$, so $\Delta U = -83 \,J$. The change in internal energy is given by $\Delta U = n C_V \Delta T$. Given $C_p = \frac{11}{10} R$, we find $C_V = C_p - R = \frac{11}{10} R - R = \frac{1}{10} R$. Substituting the values: $-83 = 1 	imes (\frac{1}{10} 	imes 8.3) 	imes (T_f - 125)$. $-83 = 0.83 	imes (T_f - 125)$. $T_f - 125 = \frac{-83}{0.83} = -100$. $T_f = 125 - 100 = 25^{\circ} C$.

An ideal gas has a specific heat capacity at constant pressure of $\frac{11}{10} R$. If one mole of this ideal gas at $125^{\circ} C$ does $83 \,J$ of work adiabatically, then the final temperature of the gas would be (Universal gas constant, $R=8.3 \,J \,K^{-1} \,mol^{-1}$ ). (in $^{\circ} C$)

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