One mole of an non-ideal gas undergoes a change of state from (2.0 atm, 3.0 L, 95 K) to (4.0 atm, 5. — Thermodynamics and Thermochemistry Chemistry Question
Question
One mole of an non-ideal gas undergoes a change of state from (2.0 atm, 3.0 L, 95 K) to (4.0 atm, 5.0 L, 245 K) with a change in internal energy, $\Delta U = 30.0\text{ L-atm}$. The enthalpy change ($\Delta H$) of the process in L-atm is
Answer: 0044
💡 Solution & Explanation
Enthalpy change $\Delta H = \Delta U + \Delta(PV)$. Here, $\Delta(PV) = P_2 V_2 - P_1 V_1 = (4.0 \times 5.0) - (2.0 \times 3.0) = 20.0 - 6.0 = 14.0\text{ L-atm}$. Thus, $\Delta H = 30.0 + 14.0 = 44.0\text{ L-atm}$. Format as 0044.
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