ThermodynamicsMedium
Question
One mole of a non-ideal gas undergoes a change of state (2 atm, 3 L, 95 K) → (4 atm, 5 L, 245 K) with a change in internal energy, ΔU = 30.0 L-atm. The change in enthalpy (ΔH) of the process in L-atm is
Options
A.40.0
B.42.3
C.44.0
D.undefined, because pressure is not constant
Solution
$\Delta H = \Delta U = \Delta(PV) = \Delta U + \left( P_{2}V_{2} - P_{1}V_{1} \right)$
$= 30.0\text{ L-atm + }(4 \times 5 - 2 \times 3)\text{ L-atm} $$$\text{=44.0 L-atm}$$
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