Question

Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

• A. $2M_A=M_B$
• B. $M_A=4M_B$
• C. $4M_A=M_B$
• D. $M_A=2M_B$

Answer 1

The correct option is C

$4M_A=M_B$

Mass of gas A , $W_A$ = 1g

Mass of gas B, $W_B$ = 2g

Pressure exerted by the gas A = 2 bar

Total pressure due to both the gases = 3 bar

In this case temperature & volume remain constant

Now if $W_A$ and $W_B$ are molar masses of the gases A and B respectively,therefore

$P_{A}V=W_A\frac{RT}{M_A}and{P}_{total}V=(\frac{W_A}{M_A}+$$\frac{W_B}{M_B})RT$

$\Rightarrow 2\times V=1\times \frac{RT}{M_A}$

and$3\times V=(\frac{1}{M_A}+\frac{2}{M_B})RT$

From these two equations, we get

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

$\frac{3}{2}=\frac{(M_B+2M_A)}{M_B}$

This result in $2\frac{M_A}{M_B}=\frac{3}{2}-1=0.5$

OR $M_B=4M_A$

Thus, a relationship between the molecular masses of A and B is given by

$4M_A=M_B$