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

Using the ideal gas law: \[ P V = n R T \] where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. 
Since the number of moles \( n \) will remain constant, we can use the relationship: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] 
From the given, we know: - \( P_1 = 8 \, \text{kPa} \), \( T_1 = 1000 \, \text{K} \), and \( V_1 = V \), - \( P_2 = 7 \, \text{kPa} \), \( T_2 = 500 \, \text{K} \), and \( V_2 = 2V \). 
At steady state, both vessels will reach a common pressure \( P_f \), and the volume of the combined system will be \( V + 2V = 3V \), with a common temperature of 600 K. 
Using the ideal gas law to find the final pressure: \[ P_f = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{(V_1 + V_2) T_f} \] 
Substituting the values: \[ P_f = \frac{8 \times 1 \times 500 + 7 \times 2 \times 1000}{(1 + 2) \times 600} = 6 \, \text{kPa} \] 
Thus, the pressure in both vessels will be 6 kPa, and the correct answer is (2).