Question

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

Answer 1

Given that,

weight of gas A = 2g

Pressure of A = 1 atm, T = 298 K

Now, another gas is introduced, then, weight of gas B = 3g

 Pressure of mixture = 1.5 atm

From, Dalton's law of partial pressure :

Answer 2

Let Partial Pressure of A be p_A and that of B ne p_B. Again let the molecular weights of A and be M_A and M_B. If V be the volume of the container then at 25℃ or T=298K we get the following relations 

 PA=2/M_A×RT

And p_BV=3/M_B×RT

pA/pB=(2MB)/(3MA)

=>MA/MB=2/3×pB/pA=2/3×(1.5-1)/1=1/3