Question

At a pressure \( P \) and temperature \( 127^\circ C \), a vessel contains 21 g of a gas. A small hole is made into the vessel so that the gas leaks out. At a pressure of \( \frac{2P}{3} \) and a temperature of \( t^\circ C \), the mass of the gas leaked out is 5 g. Then \( t \) is: 

• A. \( 273^\circ C \)
• B. \( 77^\circ C \)
• C. \( 350^\circ C \)
• D. \( 87^\circ C \)

Hint:

For gas leakage problems, apply the ideal gas law equation: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \times \frac{m_2}{m_1} \] where mass changes affect the number of moles.

Answer 1

The Correct Option is B

Step 1: Use the Ideal Gas Equation From the ideal gas equation: \[ PV = nRT \] where: - \( P \) is pressure, - \( V \) is volume, - \( n \) is the number of moles, - \( R \) is the universal gas constant, - \( T \) is temperature. Since the volume remains constant, we can use: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \times \frac{m_2}{m_1} \] where: - \( P_1 = P \), \( T_1 = 127 + 273 = 400K \), \( m_1 = 21g \), - \( P_2 = \frac{2P}{3} \), \( m_2 = 21 - 5 = 16g \), \( T_2 = t + 273 \).

 Step 2: Solve for \( t \) \[ \frac{P}{400} = \frac{\frac{2P}{3}}{T_2} \times \frac{16}{21} \] Cancel \( P \) and solving for \( T_2 \): \[ T_2 = \frac{2}{3} \times \frac{16}{21} \times 400 \] \[ T_2 = 350K \] \[ t = 350 - 273 = 77^\circ C \] Thus, the correct answer is: \[ \mathbf{77^\circ C} \]