Final Pressure of Two Connected Vessels At Different Temperatures

To find the final pressure in the connected vessels, we apply the Ideal Gas Law ($PV = nRT$) and the principle of conservation of total moles of air in the system.

1. Initial State

Let the volume of each vessel be $V$. Initially, both vessels are at a pressure $P_i = 90 \text{ kPa}$ and a temperature $T_i = 400 \text{ K}$.

The initial number of moles in vessel 1 ($n_{1i}$) and vessel 2 ($n_{2i}$) are: $n_{1i} = \frac{P_i V}{R T_i} = \frac{90 \cdot V}{R \cdot 400}$ $n_{2i} = \frac{P_i V}{R T_i} = \frac{90 \cdot V}{R \cdot 400}$

Thus, the total initial number of moles ($n_{\text{total}}$) is: $n_{\text{total}} = n_{1i} + n_{2i} = \frac{2 \times 90 \cdot V}{400 \cdot R} = \frac{180 \cdot V}{400 \cdot R}$

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2. Final State

Since the two vessels remain connected through a narrow tube, the pressure in both vessels will equalize to a final common pressure, $P_f$.

• Vessel 1 is maintained at $T_{1f} = 400 \text{ K}$.
• Vessel 2 is heated to $T_{2f} = 500 \text{ K}$.

The final number of moles in vessel 1 ($n_{1f}$) and vessel 2 ($n_{2f}$) are: $n_{1f} = \frac{P_f V}{R \cdot 400}$ $n_{2f} = \frac{P_f V}{R \cdot 500}$

The total final number of moles is: $n_{\text{total}} = n_{1f} + n_{2f} = \frac{P_f V}{R} \left( \frac{1}{400} + \frac{1}{500} \right)$

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3. Conservation of Moles

Since the system is closed, the total number of moles remains constant: $\frac{P_f V}{R} \left( \frac{1}{400} + \frac{1}{500} \right) = \frac{180 \cdot V}{400 \cdot R}$

Dividing both sides by $\frac{V}{R}$: $P_f \left( \frac{5 + 4}{2000} \right) = \frac{180}{400}$

$P_f \left( \frac{9}{2000} \right) = \frac{9}{20}$

Solving for $P_f$: $P_f = \frac{9}{20} \times \frac{2000}{9} = 100 \text{ kPa}$

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Conclusion

The final pressure in the vessels is $100 \text{ kPa}$.

Correct Option: A