Question

Two separate bulbs contain ideal gases A and B. The density of A is twice that of gas B. The molecular mass of gas A is half that of B. If two gases are at the same temperature, the ratio of the pressure of A to that of B is

• A. $2$
• B. $\frac{1}{2}$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

The correct option is C

$4$

Step 1: Given data

$$\begin{gathered} M_A=\frac{1}{2}{M}_B \\ {\rho }_A=2{\rho }_B \end{gathered}$$

Here,

$M_A$ is the mass of the gas A.

$M_B$ is the mass of the gas B.

${\rho }_A$ is the density of gas A.

${\rho }_B$ is the density of gas B.

Step 2: Equation of pressure of ideal gas

The equation for the pressure of ideal gas can be written as

$PM=\rho RT$

Here, $P$ is the pressure, $M$ is the molecular mass, $\rho$ is the density, $R$ is the gas constant, and $T$ is the temperature.

Step 3: Equation of pressure of the ideal gas in the bulb A

$P_A{M}_A={\rho }_{A}R{T}_A$

Step 4: Equation of pressure of the ideal gas in the bulb B

$P_B{M}_B={\rho }_{B}R{T}_B$

The density of A is twice that of gas B, the molecular mass of gas A is half that of B, and the temperatures are the same.

The ratio of pressures can be written as follows:

$$\begin{gathered} \frac{P_A{M}_A}{P_B{M}_B}=\frac{{\rho }_{A}R{T}_A}{{\rho }_{B}R{T}_B} \\ \frac{P_A\times {\displaystyle \frac{1}{2}}{M}_B}{P_B{M}_B}=\frac{2{\rho }_{B}R{T}_B}{{\rho }_{B}R{T}_B} \\ \frac{P_A}{P_B}=4 \end{gathered}$$

Hence, the ratio of the pressure of A to that of B is $4$.

Therefore, option (C) is correct.