Question

An ideal gas expands by performing 200 J of work, during this internal energy increases by 432 J. What is enthalpy change?

• A. 200 J
• B. 232 J
• C. 432 J
• D. 632 J

Hint:

Logic Tip: Enthalpy change ($\Delta H$) represents the total heat transferred at constant pressure ($q_p$). Using the first law $\Delta U = q + w$, where $w = -200 \text{ J}$ (expansion work is lost energy) and $\Delta U = +432 \text{ J}$, we get $432 = q_p - 200 \implies q_p = 632 \text{ J}$. Since $\Delta H = q_p$, $\Delta H = 632 \text{ J}$.

Answer 1

The Correct Option is D

Concept:
The change in enthalpy ($\Delta H$) at constant pressure is related to the change in internal energy ($\Delta U$) and the pressure-volume work ($P\Delta V$) by the definition of enthalpy ($H = U + PV$). For a process at constant pressure: $$\Delta H = \Delta U + P\Delta V$$ The work done \textit{by} the gas during expansion is $W_{\text{expansion = P\Delta V$.
Step 1: Identify the given thermodynamic parameters.
The gas expands by performing work. The magnitude of work done by the gas is $200 \text{ J}$. Therefore, the pressure-volume work term is: $$P\Delta V = 200 \text{ J}$$ (Note: Using IUPAC sign convention, work done \textit{by} the system is negative, so $w = -200 \text{ J}$. Since $w = -P\Delta V$, we still have $P\Delta V = 200 \text{ J}$). The internal energy increases by $432 \text{ J}$. $$\Delta U = +432 \text{ J}$$
Step 2: Calculate the enthalpy change ($\Delta H$).
Substitute the values into the enthalpy equation: $$\Delta H = \Delta U + P\Delta V$$ $$\Delta H = 432 \text{ J} + 200 \text{ J}$$ $$\Delta H = 632 \text{ J}$$