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A certain mass of ideal gas is taken from state (1 L, 1 atm, 300 K) to state (4 L, 5 atm) against a constant external pressure of 1 atm. The molar heat capacity of the process is 21 J/degC. Calculate the enthalpy change (delta_H) for the process. (Given: 1 L atm = 0.1 kJ)

1. 119.7 kJ
2. 121.3 kJ
3. 123.0 kJ
4. 118.0 kJ

Correct answer: 121.3 kJ

Solution

From ideal gas law (nR is constant across states): T2/T1 = (P2V2)/(P1V1) = (5*4)/(1*1) = 20, so T2 = 300*20 = 6000 K, delta_T = 5700 K. Heat absorbed: Q = C * delta_T = 21 * 5700 = 119700 J = 119.7 kJ. Work done by gas: W = P_ext * delta_V = 1 * 3 = 3 L atm = 0.3 kJ. By first law: delta_U = Q - W = 119.7 - 0.3 = 119.4 kJ. Change in PV: delta(PV) = (P2V2 - P1V1) = (5*4 - 1*1) * 0.1 kJ = 19 * 0.1 = 1.9 kJ. Therefore delta_H = delta_U + delta(PV) = 119.4 + 1.9 = 121.3 kJ.

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