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).
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- 4.4
- 6
- 24
- 18
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to determine the final pressure in the vessels when they are connected and temperature is stabilized. Let's break down the problem using the ideal gas law.
The ideal gas law is given by the formula:
\(PV = nRT\)
where:
- \(P\) is the pressure of the gas.
- \(V\) is the volume of the gas.
- \(n\) is the number of moles of the gas.
- \(R\) is the ideal gas constant.
- \(T\) is the temperature of the gas in Kelvin.
Given data:
- Volume of the larger vessel, \(V_1\), is twice that of the smaller vessel, \(V_2\). So, \(V_1 = 2V_2\).
- Pressure in larger vessel, \(P_1 = 8\) kPa, and \(T_1 = 1000\) K.
- Pressure in smaller vessel, \(P_2 = 7\) kPa, and \(T_2 = 500\) K.
Steps to determine the final pressure:
- Calculate the moles of gas in each vessel using the ideal gas equation:
- \(n_1 = \frac{P_1 V_1}{R T_1}\)
- \(n_2 = \frac{P_2 V_2}{R T_2}\)
- Since \(V_1 = 2V_2\), we substitute to find:
- \(n_1 = \frac{8 \times 2V_2}{RT \times 1000}\)
- \(n_2 = \frac{7 \times V_2}{RT \times 500}\)
- This simplifies to:
- \(n_1 = \frac{16V_2}{RT \times 1000}\)
- \(n_2 = \frac{7V_2}{RT \times 500}\) which simplifies to \(n_2 = \frac{14V_2}{RT \times 1000}\)
- Total moles \(n_{\text{total}} = n_1 + n_2\)
- Now, connecting and maintaining them at 600 K, use:
- \(P_{\text{final}} \cdot 3V_2 = n_{\text{total}} \cdot RT_{\text{final}}\)
- Substitute the values:
- \(P_{\text{final}} \cdot 3V_2 = \left(\frac{16V_2}{RT \times 1000} + \frac{14V_2}{RT \times 1000}\right) \cdot R \times 600\)
- Simplifying gives:
- \(P_{\text{final}} \cdot 3V_2 = \left(\frac{30V_2}{RT \times 1000}\right) \cdot 600\)
- Further simplification:
- \(P_{\text{final}} = \frac{30 \times 600}{1000 \times 3}\)
- This yields \(P_{\text{final}} = 6\) kPa.
Conclusion: The final pressure in the connected vessels, when stabilized at 600 K, is 6 kPa.
Approach Solution -2
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).
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