Question

Pressure of $1 \,g$ of an ideal gas $A$ at $27 \, ^{\circ}C$ is found to be $2$ bar. When $2 \,g$ of another ideal gas $B$ is introduced in the same flask at same temperature the pressure becomes $3$ bar. What would be the ratio of their molecular masses?

• A. $4 : 1$
• B. $1 : 4$
• C. $1 : 8$
• D. $2 : 8$

Answer 1

Answer: (B)

$1 : 4$

Answer 2

For gas $A$, $P_{A} V$ =$\frac{m_{A}}{M_{A}} RT$ $\quad$ $\left(1\right)$ For gas $B$, $P_{B}V$ =$\frac{m_{B}}{M_{B}}RT$ $\quad$ $\left(2\right)$ Dividing equation $\left(1\right)$ by equation $\left(2\right)$ gives $\frac{P_{A}}{P_{B}}$=$\frac{m_{A}}{m_{B}} \frac{M_{B}}{M_{A}}$ $P_{A}+P_{B}=3$ $\Rightarrow$ $P_{B}$ = $3-2=1$ bar $\frac{M_{A}}{M_{B}}$ = $\left(\frac{m_{A}}{m_{B}}\right)$ $\left(\frac{P_{B}}{P_{A}}\right)$ =$\left(\frac{1g}{2 g}\right)$ $\left(\frac{1 bar}{2 bar}\right)$ $\Rightarrow$ $\frac{M_{A}}{M_{B}}$ = $1 : 4$