Solution
System Analysis:
We are dealing with a mono-atomic ideal gas ($\gamma = 5/3$). The process involves heating the gas to push liquid out of a U-tube. The expansion is slow (quasi-static).
1. Geometry and Initial State:
- Total volume of liquid = $V_0$.
- Initial volume of gas in the flask $V_{gas, i} = V_0$.
- Initial pressure $P_i = p_0$ (Atmospheric).
- Since the U-tube cross-section is $S$ and total liquid volume is $V_0$, the length of the liquid column is $L_{total} = V_0/S$. Initially, the liquid height in each arm is equal, so $h_i = \frac{V_0}{2S}$.
2. Final State Analysis:
We are driving half of the liquid out. Since the initial liquid volume was $V_0$, the remaining liquid volume is $V_0/2$.
The gas pushes the liquid level down in the right arm. Simultaneously, liquid rises in the left arm and spills over. Since the final volume inside the tube is exactly half the original, and the original height $h_i$ corresponded to volume $V_0/2$ per arm, this implies the right arm is now completely empty of liquid (gas reaches the bottom bend), and the left arm is full (liquid at the top).
- Height difference between the two liquid levels: $y_{final} = h_i = \frac{V_0}{2S}$. Let’s call this $H$.
- Final Gas Volume $V_f = V_0 (\text{flask}) + S \cdot H (\text{right arm}) = V_0 + S(\frac{V_0}{2S}) = \frac{3V_0}{2}$.
- Final Pressure $P_f$: The gas balances atmospheric pressure plus the liquid column of height $H$. $$P_f = p_0 + \rho g H = p_0 + \rho g \left( \frac{V_0}{2S} \right)$$
3. Work Done by Gas ($W$):
The work done by the gas consists of pushing against the atmospheric pressure and increasing the potential energy of the liquid column. We can calculate this by integrating $P \, dV$.
Let $y$ be the instantaneous depth of the meniscus in the right arm relative to the top of the left arm (where liquid is spilling). The pressure at any instant is:
$$P(y) = p_0 + \rho g y$$The gas expands by pushing the meniscus down in the right arm. The change in volume is $dV = S \, dy$. The limits of integration for $y$ are from $0$ to $H$ (where $H = \frac{V_0}{2S}$).
$$W = \int_{0}^{H} P(y) \, dV = \int_{0}^{H} (p_0 + \rho g y) S \, dy$$ $$W = S \left[ p_0 y + \frac{1}{2}\rho g y^2 \right]_0^H$$ $$W = S \left( p_0 H + \frac{1}{2}\rho g H^2 \right)$$Substituting $H = \frac{V_0}{2S}$:
$$W = S \left( p_0 \frac{V_0}{2S} + \frac{1}{2}\rho g \frac{V_0^2}{4S^2} \right) = \frac{p_0 V_0}{2} + \frac{\rho g V_0^2}{8S}$$4. Change in Internal Energy ($\Delta U$):
For a mono-atomic gas, $\Delta U = \frac{3}{2}(P_f V_f – P_i V_i)$.
$$P_i V_i = p_0 V_0$$ $$P_f V_f = \left( p_0 + \frac{\rho g V_0}{2S} \right) \left( \frac{3V_0}{2} \right) = \frac{3 p_0 V_0}{2} + \frac{3 \rho g V_0^2}{4S}$$ $$\Delta(PV) = P_f V_f – P_i V_i = \frac{1}{2} p_0 V_0 + \frac{3 \rho g V_0^2}{4S}$$ $$\Delta U = \frac{3}{2} \left[ \frac{p_0 V_0}{2} + \frac{3 \rho g V_0^2}{4S} \right] = \frac{3 p_0 V_0}{4} + \frac{9 \rho g V_0^2}{8S}$$5. Total Heat Supplied ($Q$):
According to the First Law of Thermodynamics, $Q = \Delta U + W$.
$$Q = \left( \frac{3 p_0 V_0}{4} + \frac{9 \rho g V_0^2}{8S} \right) + \left( \frac{p_0 V_0}{2} + \frac{\rho g V_0^2}{8S} \right)$$Group the terms with $p_0 V_0$ and $\rho g V_0^2/S$:
$$Q = p_0 V_0 \left( \frac{3}{4} + \frac{2}{4} \right) + \frac{\rho g V_0^2}{S} \left( \frac{9}{8} + \frac{1}{8} \right)$$ $$Q = \frac{5}{4} p_0 V_0 + \frac{10}{8} \frac{\rho g V_0^2}{S} = \frac{5}{4} p_0 V_0 + \frac{5}{4} \frac{\rho g V_0^2}{S}$$Factoring out $\frac{5 p_0 V_0}{4}$: