Question

There are two vessels filled with an ideal gas where volume of one is double the volume of the other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).

• A. 4.4
• B. 6
• C. 24
• D. 18

Hint:

For connected vessels containing an ideal gas, the pressure is determined by balancing the total mass of gas and its temperature across the vessels.

Answer 1

The Correct Option is B

To determine the final pressure after connecting the vessels and stabilizing the temperature, we will apply the ideal gas law.

The ideal gas law is expressed as:

\(PV = nRT\)

where:

• \(P\) represents gas pressure.
• \(V\) represents gas volume.
• \(n\) denotes the number of moles of gas.
• \(R\) is the ideal gas constant.
• \(T\) is the gas temperature in Kelvin.

Provided Data:

• The volume of the larger vessel, \(V_1\), is double that of the smaller vessel, \(V_2\), thus \(V_1 = 2V_2\).
• For the larger vessel: \(P_1 = 8\) kPa and \(T_1 = 1000\) K.
• For the smaller vessel: \(P_2 = 7\) kPa and \(T_2 = 500\) K.

Procedure to Calculate Final Pressure:

1. Calculate the moles of gas in each vessel using the ideal gas equation:
   • \(n_1 = \frac{P_1 V_1}{R T_1}\)
   • \(n_2 = \frac{P_2 V_2}{R T_2}\)
2. Substitute \(V_1 = 2V_2\):
   • \(n_1 = \frac{8 \times 2V_2}{R \times 1000}\)
   • \(n_2 = \frac{7 \times V_2}{R \times 500}\)
3. Simplify the mole calculations:
   • \(n_1 = \frac{16V_2}{1000R}\)
   • \(n_2 = \frac{7V_2}{500R} = \frac{14V_2}{1000R}\)
4. Calculate the total moles: \(n_{\text{total}} = n_1 + n_2\)
5. With the vessels connected and stabilized at 600 K, apply the ideal gas law:
   • \(P_{\text{final}} \cdot (V_1 + V_2) = n_{\text{total}} \cdot R T_{\text{final}}\)
   • Since \(V_1 = 2V_2\), the total volume is \(3V_2\).
   • \(P_{\text{final}} \cdot 3V_2 = n_{\text{total}} \cdot R \times 600\)
6. Substitute the derived mole values:
   • \(P_{\text{final}} \cdot 3V_2 = \left(\frac{16V_2}{1000R} + \frac{14V_2}{1000R}\right) \cdot 600R\)
7. Simplify the equation:
   • \(P_{\text{final}} \cdot 3V_2 = \left(\frac{30V_2}{1000R}\right) \cdot 600R\)
8. Further simplification to find \(P_{\text{final}}\):
   • \(P_{\text{final}} = \frac{30 \times 600}{1000 \times 3}\)
9. The result is: \(P_{\text{final}} = 6\) kPa.

Conclusion: The final pressure within the connected vessels, after reaching a stable temperature of 600 K, is 6 kPa.