Q168 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
Two identical vessels are completely filled with two ideal gases A and B. If the ratio of mass of gas A to that of gas B is 1:2 and the ratio of rms speed of molecules of gas A to that of gas B is 2:1, the ratio of pressure exerted by molecules of gas B, on the walls of vessel to that by molecules of gas A is
- A$4 : 1$
- B$2 : 1$
- Correct answerC$1 : 2$
- D$1 : 4$
Solution
The pressure exerted by an ideal gas is given by $P = \dfrac{1}{3} \rho v_{rms}^2 = \dfrac{1}{3} \dfrac{m}{V} v_{rms}^2$
Since the vessels are identical, the volume $V$ is the same for both gases.
Therefore, $P \propto m v_{rms}^2$
The ratio of pressures is given by:
$\dfrac{P_B}{P_A} = \dfrac{m_B}{m_A} \times \left( \dfrac{v_{rms, B}}{v_{rms, A}} \right)^2$
Given $\dfrac{m_A}{m_B} = \dfrac{1}{2} \Rightarrow \dfrac{m_B}{m_A} = 2$
And $\dfrac{v_{rms, A}}{v_{rms, B}} = \dfrac{2}{1} \Rightarrow \dfrac{v_{rms, B}}{v_{rms, A}} = \dfrac{1}{2}$
Substituting the values:
$\dfrac{P_B}{P_A} = 2 \times \left( \dfrac{1}{2} \right)^2 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$
The ratio of pressure exerted by gas B to that by gas A is $1 : 2$.
Answer: $1 : 2$
Since the vessels are identical, the volume $V$ is the same for both gases.
Therefore, $P \propto m v_{rms}^2$
The ratio of pressures is given by:
$\dfrac{P_B}{P_A} = \dfrac{m_B}{m_A} \times \left( \dfrac{v_{rms, B}}{v_{rms, A}} \right)^2$
Given $\dfrac{m_A}{m_B} = \dfrac{1}{2} \Rightarrow \dfrac{m_B}{m_A} = 2$
And $\dfrac{v_{rms, A}}{v_{rms, B}} = \dfrac{2}{1} \Rightarrow \dfrac{v_{rms, B}}{v_{rms, A}} = \dfrac{1}{2}$
Substituting the values:
$\dfrac{P_B}{P_A} = 2 \times \left( \dfrac{1}{2} \right)^2 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$
The ratio of pressure exerted by gas B to that by gas A is $1 : 2$.
Answer: $1 : 2$
Hence, the correct answer is option: 'C'.