One mole of non-ideal gas undergoes a change of state from (1atm, 3L, 200K) to (4 atm, 5L, 250K) with a change in internal energy $(ΔU) = 40 L . atm$ . the change in enthalpy of the process in L.atm.

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answer is 57.

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Detailed Solution

When both P and V are changing
$\begin{array}{l} ΔH = ΔU + Δ (PV) \\ = ΔU + (P_{2} V_{2} - P_{1} V_{1}) \\ ΔH = 40 + (20 - 3) \Rightarrow 57 L . atm \end{array}$

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