One mole of non-ideal gas undergoes a change of state (1.0 atm,3.0 L, 200K) to (4.0 atm, 5.0 L, 200K) a change in internal energy $(△ U)$ = 40 L-atm. The change in enthalpy of the process in L-atm:

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None of these

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Solution

The correct option is C None of these
Solution:- (D) None of these
Given:-
$(1 atm, 3.0 L, 200 K) \to (4 atm, 5.0 L, 200 K)$
Therefore,
$\begin{matrix} P_{1} = 1 a t m & P_{2} = 4 a t m V_{1} = 3 L & V_{2} = 5 L T_{1} = 200 K & T_{2} = 200 K \end{matrix}$
$Δ U = 40 L - a t m$
As the temperature is constant, thus enthalpy change will be zero, i.e.,
$Δ H = 0$
Hence the enthalpy change is $0$ .

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