Question

A closed vessel contains a gas at a pressure $P$. If 50% of the mass of the gas is removed and rms speed of the gas molecules is increased by 20%, then the pressure of the remaining gas is

• A. $\frac{16P}{25}$
• B. $\frac{8P}{25}$
• C. $\frac{9P}{25}$
• D. $\frac{18P}{25}$

Hint:

The pressure of a gas depends on both the mass (or number of molecules) and the rms speed, which is related to temperature via $v_{{rms}} \propto \sqrt{T}$.

Answer 1

The Correct Option is D

For an ideal gas, pressure is given by \( P = \frac{1}{3} \frac{m}{V} v_{\text{rms}}^2 \), where \( m \) is the total mass of the gas, \( V \) is the volume, and \( v_{\text{rms}} \) is the rms speed of the molecules. Since \( m = nM \) (where \( n \) is the number of moles and \( M \) is the molar mass), we can write \( P \propto \frac{m}{V} v_{\text{rms}}^2 \).

Initially, \( P_1 = P \), mass \( m_1 = m \), and rms speed \( v_{\text{rms1}} = v \). After removing 50% of the mass, \( m_2 = 0.5m \). The rms speed increases by 20%, so \( v_{\text{rms2}} = 1.2v \). Since the vessel is closed, \( V \) remains constant. The new pressure \( P_2 \) is:
\[ P_2 \propto \frac{m_2}{V} v_{\text{rms2}}^2 \quad \text{and} \quad P_1 \propto \frac{m_1}{V} v_{\text{rms1}}^2 \]
\[ \frac{P_2}{P_1} = \frac{m_2}{m_1} \left(\frac{v_{\text{rms2}}}{v_{\text{rms1}}}\right)^2 = (0.5) \times (1.2)^2 = 0.5 \times 1.44 = 0.72 \]
\[ P_2 = 0.72P = \frac{72}{100}P = \frac{18}{25}P \]
The new pressure is \( \frac{18P}{25} \).