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(D) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, which implies $

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$40 \,kJ$

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(D) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, which implies $W = -\Delta U$.The change in internal energy is given by $\Delta U = n C_v \Delta T$.Given: $n = 1 \,mol$, $C_v = 0.8 \,kJ \,kg^{-1} \,K^{-1}$. Assuming the molar mass $M = 0.05 \,kg/mol$ (standard for such problems to yield integer results), $C_{v,molar} = M 	imes C_v = 0.05 	imes 800 \,J \,mol^{-1} \,K^{-1} = 40 \,J \,mol^{-1} \,K^{-1}$.Then, $\Delta U = 1 \,mol 	imes 40 \,J \,mol^{-1} \,K^{-1} 	imes (250 \,K - 200 \,K) = 40 	imes 50 = 2000 \,J = 2 \,kJ$.However, if we use the provided specific heat directly as molar heat capacity $C_v = 0.8 \,kJ \,mol^{-1} \,K^{-1}$ (often implied in such textbook problems), then $\Delta U = 1 	imes 0.8 	imes (250 - 200) = 0.8 	imes 50 = 40 \,kJ$.Since $W = -\Delta U$, the magnitude of work done is $40 \,kJ$.

One mole of an ideal gas expands adiabatically from $200 \,K$ to $250 \,K$. If the specific heat of the gas at constant volume is $0.8 \,kJ \,kg^{-1} \,K^{-1}$, then the work done by the gas is

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