The increase in Gibbs free energy (in kJ) of 13 g of ethanol (density ), when the pressure is increa — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$dG = VdP - SdT$. For an isothermal process, $\Delta G = \int VdP$. Assuming the volume of the liquid is relatively incompressible and constant: $V = \frac{\text{mass}}{\text{density}} = \frac{13\text{ g}}{0.78\text{ g/cm}^3} = 16.667\text{ cm}^3 = 16.667 \times 10^{-6}\text{ m}^3$. $\Delta P = 3001 - 1 = 3000\text{ bar} = 3 \times 10^8\text{ Pa}$. $\Delta G = (16.667 \times 10^{-6}) \times (3 \times 10^8) = 5000\text{ J} = 5\text{ kJ}$.