Question

$2g$ of gas $A$ introduced in an evacuated flask at ${25}^{\circ }C$. The pressure of the gas is $1atm$. Now, $3g$ of another gas $B$ is introduced in the same flask so total pressure becomes $1.5atm$. The ratio of the molecular mass of $A$ and $B$ is:

• A. $\frac{3}{1}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

Ideal Gas Equation accounts for the changes in Pressure, Volume or Temperature of any gas.

$PV=nRT-------\left(1\right)$

where,

R is Universal Gas Constant

n is mole

GAS A

$P_A=1atm$

$T_A={25}^{o}C$

$V_A=V_o$

$n_A=\frac{2}{M_A}$

GAS B + GAS A

$P_{A+B}=1.5atm$

$T_{A+B}=T_A$

$V_{A+B}=V_A$

$n_{A+B}=\frac{2}{M_A}+\frac{3}{M_B}$

from equation (1)

$\frac{P_A}{n_B}=\frac{P_{A+B}}{n_{A+B}}$

$\frac{1}{\frac{2}{M_A}}=\frac{1.5}{\frac{2}{M_A}+\frac{3}{M_B}}$

$M_A=\frac{3M_A{M}_B}{2M_B+3M_A}$

$2M_B+3M_A=3M_A$

$3M_A=M_B$

$\frac{M_A}{M_B}=\frac{1}{3}$

Option B is correct answer