Question

A vessel contains 8 g of air at 400 K. Some amount of air leaks out through the hole provided to it. After some time, pressure is halved and temperature is changed to 300 K. Find the mass of the air escaped.

• A. \(5.33 { g}\)
• B. \(2.67 { g}\)
• C. \(6 { g}\)
• D. \(3.27 { g}\)

Hint:

Remember to check units for temperature (K) and pressure (assumed constant here) in gas law calculations.

Answer 1

The Correct Option is B

Step 1: Use the ideal gas law. Initially, let \( n_i \) be the number of moles of air, \( V \) the volume, and \( R \) the ideal gas constant: \[ PV = n_iRT_i \quad {and} \quad \frac{P}{2}V = n_fR(300 { K}), \] where \( n_i = \frac{8}{29} \) moles (assuming air is mostly nitrogen, \( M = 29 { g/mol} \)), and \( T_i = 400 { K} \). 
Step 2: Calculate the final number of moles and the difference. From the equations, it follows that: \[ n_f = \frac{n_i T_i}{2 \times 300} = \frac{\frac{8}{29} \times 400}{600} = \frac{8}{43.5} { moles}. \] The moles of air escaped: \[ \Delta n = n_i - n_f = \frac{8}{29} - \frac{8}{43.5} \approx 0.092 { moles}. \] The mass of the air escaped: \[ \Delta m = \Delta n \times 29 \approx 2.67 { g}. \]