Question

When 2g of a gas A are introduced into an evacuated flask kept at ${25}^{0}C$, the pressure is found to be one atmosphere, if 3g of another gas B are then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, Calculate the ratio of molecular weights $M_A:M_B$.

Answer 1

When 2 grams of a gas A is introduced in an evacuated flask kept at ${25}^{o}C$; the pressure was found to be 1 atm .i.e.

Partial Pressure of gas A = 1 atm.

And let moles be $n_A=\frac{2}{M_A}$

if 3 g of another gas B is then added to the same flask;

Partial Pressure of gas B = 1.5 atm – 1 atm. = 0.5 atm

moles of gas B $=\frac{3}{M_B}$

Again the temp and volume are constant therefore,

$\frac{P_1}{n_1}=\frac{P_2}{n_2}$

Hence, $\frac{1}{\frac{2}{M_A}}=\frac{0.5}{\frac{3}{M_B}}$

$M_A=\frac{0.5M_B}{3}\times 2$

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

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

or

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