As 1 mole of liquid water is heated from 288 to 298 K, it expands slightly against the atmosphere. G — Thermodynamics and Thermochemistry Chemistry Question
Question
As 1 mole of liquid water is heated from 288 to 298 K, it expands slightly against the atmosphere. Given coefficient of cubic expansion of water is $0.0002/^\circ\text{C}$; density of water $= 1.0\text{ g cm}^{-3}$ and external pressure $= 1\text{ bar}$. If the magnitude of work for this process is $X\text{ J}$, then the value of $\frac{X}{6 \times 10^{-4}}$ is
💡 Solution & Explanation
Mass of 1 mole water is 18 g. Initial volume $V_o = 18\text{ g} / 1.0\text{ g/cm}^3 = 18\text{ cm}^3 = 18 \times 10^{-6}\text{ m}^3$. Change in volume $\Delta V = V_o \gamma \Delta T = (18 \times 10^{-6}) \times 0.0002 \times 10 = 36 \times 10^{-9}\text{ m}^3$. Work done $W = P_{\text{ext}} \Delta V = 10^5\text{ Pa} \times 36 \times 10^{-9}\text{ m}^3 = 3.6 \times 10^{-3}\text{ J}$. Thus $X = 3.6 \times 10^{-3}$. The required ratio is $\frac{3.6 \times 10^{-3}}{6 \times 10^{-4}} = \frac{36}{6} = 6$.