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(C) For an isothermal process,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q + W$,so $Q = -W$

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$200$

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(C) For an isothermal process,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q + W$,so $Q = -W$.Since the process is irreversible and occurs against a constant external pressure $P_{ext}$,the work done is $W = -P_{ext} 	imes \Delta V$.Given $P_{ext} = 5 \ atm$,$V_1 = 60 \ L$,and $V_2 = 20 \ L$,the change in volume $\Delta V = V_2 - V_1 = 20 \ L - 60 \ L = -40 \ L$.$W = -5 \ atm 	imes (-40 \ L) = 200 \ L \cdot atm$.Therefore,$Q = -W = -200 \ L \cdot atm$.

Three moles of an ideal gas are compressed isothermally from $60 \ L$ to $20 \ L$ using a constant external pressure of $5 \ atm$. The heat exchange $Q$ for the compression is $....... \ L \cdot atm$.

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