Question

Two grams of gas $A$ are introduced in an evacuated flask at ${25}^{o}C$ . The pressure of the gas is $1atm$. Now $3g$ of another gas is introduced in the same flask, the total pressure becomes $1.5atm$. The ratio of molecular mass $M_A$ and $M_B$ is $x:y$. Find the value of $x\times y$

Answer 1

For gas A : Its mass = $2g$
$P_A=1atm,T=298K$
For gas B : Its mass =$3g$
$P_B=0.5atm,T=298K$
According to Dalton's law of partial pressure
$P=P_A+P_B$
$1.5=1+P_B$
$\therefore ,P_B=0.5atm$
Now,
$PV=nRT$
$PV=\frac{w}{m}RT$
For gas A
$P_A=1atm,m=M_A,w=2g$
$1\times V=\frac{2\times RT}{M_A}...\left(i\right)$
For gas B
$P_B=0.5atm,m=M_B,w=3g$
$0.5\times V=\frac{3\times RT}{M_B}...\left(ii\right)$
Dividing equation (i) and (ii)
$0.5=\frac{3\times M_A}{2M_B}$
or
$\frac{M_A}{M_B}=\frac{1}{3}$
Hence, $1\times 3=3$.