The vapour pressure of water is 0.04 atm at . The free energy change for the following process is — Thermodynamics and Thermochemistry Chemistry Question
Question
The vapour pressure of water is 0.04 atm at $27^\circ\text{C}$. The free energy change for the following process is $\text{H}_2\text{O (g, 0.04 atm, } 27^\circ\text{C)} \to \text{H}_2\text{O (l, 0.04 atm, } 27^\circ\text{C)}$
Answer: A
💡 Solution & Explanation
At the exact specified pressure (0.04 atm) and temperature ($27^\circ\text{C}$), the liquid and gaseous phases of water reside in dynamic physical equilibrium. Because the system is at equilibrium, the change in Gibbs free energy ($\Delta G$) evaluates inherently to zero.
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