(A)→(P), (R), (T) We have PdV +VdP = nRdT Since dP = 0 and dV is negative, dT also negative. The internal energy is a function of temperature alone; hence, the internal energy decreases. Also, dU = dQ – W Since dU is negative, dQ also negative and the work is done on the gas. (B)→(P), (R) Here also, dT is negative and dV = 0. Therefore, the internal energy of C decreases and hence the heat is lost by the gas. (C)→(Q), (S) Here, dV and dT are positive. Hence, the system gains heat, and the internal energy of the system increases. (D)→(R), (T) The work is done on the gas and therefore the system loses heat.