THERMAL O29

Solution for Question 29

Step 1: Identify Initial and Final States

• Initial State:
  Volume $V_i = A h$ (from diagram).
  Pressure $P_i = p_0 + \frac{Mg}{A} + \rho g h$.
  Temperature $T_i = T_0$.
• Final State: (Liquid completely removed, piston at top)
  Volume $V_f = A (2h) = 2 V_i$ (since piston moved up by h).
  Pressure $P_f = p_0 + \frac{Mg}{A}$.
  Temperature $T_f = ?$

Step 2: Apply Ideal Gas Law

Using $\frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}$:

$$T_f = T_i \left( \frac{P_f}{P_i} \right) \left( \frac{V_f}{V_i} \right)$$

Substitute $V_f = 2V_i$:

$$T_f = T_0 \left( \frac{P_f}{P_i} \right) (2) = 2 T_0 \frac{P_f}{P_i}$$

Substitute the pressure expressions:

$$T_f = 2 T_0 \frac{p_0 + \frac{Mg}{A}}{p_0 + \frac{Mg}{A} + \rho g h}$$

Multiplying numerator and denominator by $A$:

$$T_f = 2 T_0 \left[ \frac{p_0 A + Mg}{p_0 A + Mg + \rho g h A} \right]$$

(Note: The option uses $p_A$ to denote atmospheric pressure $p_0$).