tHERMAL O14

Solution for Question 14

Step 1: Initial State

Gas is filled in flask B at pressure $P$. Flask A is evacuated (Pressure = 0). Valve C is closed.

• Initial Pressure ($P_i$) = $P$
• Initial Volume ($V_i$) = $V_B$

Step 2: Final State

Valve C is opened. The gas expands to fill both flasks A and B. The temperature remains constant (conducting materials).

• Final Volume ($V_f$) = $V_A + V_B$
• Final Pressure ($P_f$) = $P – \Delta p$ (Since pressure drops by $\Delta p$)

Calculation:

Apply Boyle’s Law ($P_i V_i = P_f V_f$):

$$P V_B = (P – \Delta p)(V_A + V_B)$$

$$P V_B = (P – \Delta p)V_A + (P – \Delta p)V_B$$

Rearranging terms to isolate $V_A$:

$$P V_B – (P – \Delta p)V_B = (P – \Delta p)V_A$$

$$V_B [ P – (P – \Delta p) ] = (P – \Delta p)V_A$$

$$V_B [ \Delta p ] = (P – \Delta p)V_A$$

$$V_A = \frac{\Delta p V_B}{P – \Delta p}$$