An amount of 5 mole at and 1 atm is converted into at and 5 atm. for the process is — Thermodynamics and Thermochemistry Chemistry Question
Question
An amount of 5 mole $\text{H}_2\text{O(l)}$ at $100^\circ\text{C}$ and 1 atm is converted into $\text{H}_2\text{O(g)}$ at $100^\circ\text{C}$ and 5 atm. $\Delta G$ for the process is
Answer: C
💡 Solution & Explanation
Breaking this process into reversible steps: (1) Vaporization at 1 atm, $100^\circ\text{C}$ gives $\Delta G = 0$. (2) Isothermal compression of 5 moles of gas from 1 atm to 5 atm. $\Delta G = nRT \ln(P_2/P_1) = 5 \times 2 \times 373 \times \ln(5/1) = 3730 \ln 5 \text{ cal}$. Total $\Delta G = 3730 \ln 5 \text{ cal}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes