Solution 32
Physical Process:
The water is at freezing point $T_0$. The gas is at $T < T_0$. Heat flows from the water to the gas. 1. The gas warms up from $T$ to $T_0$ (Isobaric expansion at $P_0$). 2. The water releases heat and freezes into ice at $T_0$. 3. Ice is less dense than water, so the volume of the water/ice layer increases.
Heat Balance:
Heat gained by gas = Latent heat released by freezing water. $$Q = n C_p (T_0 – T) = m_{ice} L$$ $$m_{ice} = \frac{n C_p (T_0 – T)}{L}$$
Displacement Calculation:
Total displacement volume $\Delta V_{total} = A \cdot y$. This is the sum of gas expansion and ice expansion.
1. Gas Expansion: $$\Delta V_{gas} = \frac{nR(T_0 – T)}{P_0}$$
2. Ice Expansion: $$\Delta V_{ice} = V_{ice} – V_{water} = m_{ice} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) = m_{ice} \frac{\rho_w – \rho_i}{\rho_i \rho_w}$$ Substituting $m_{ice}$ (with $C_p = \frac{5}{2}R$ for monoatomic gas): $$\Delta V_{ice} = \frac{n \frac{5}{2} R (T_0 – T)}{L} \left( \frac{\rho_w – \rho_i}{\rho_i \rho_w} \right)$$
Total Displacement: $$A y = \frac{nR(T_0 – T)}{P_0} + \frac{5nR(T_0 – T)}{2L} \frac{\rho_w – \rho_i}{\rho_i \rho_w}$$ $$y = \frac{nR(T_0 – T)}{A P_0} \left[ 1 + \frac{P_0}{nR(T_0 – T)} \cdot \frac{5nR(T_0 – T)}{2L} \frac{\rho_w – \rho_i}{\rho_i \rho_w} \right]$$ $$y = \frac{nR(T_0 – T)}{A P_0} \left[ 1 + \frac{5 P_0 (\rho_w – \rho_i)}{2 L \rho_w \rho_i} \right]$$