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 same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses. 

Answer 1

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 M_A & M_B are molar masses of the gases A & B respectively,therefore

p_A V= W_A RT/M_A & P_total V = (W_A/M_A + W_B/ M_B) RT

= 2 X V = 1 X RT/M_A & 3 X V = (1/M_A + 2/M_B)RT

From these two equations, we get

3/2 = (1/M_A + 2/M_B) / (1 /M_A) = (M_B + 2M_A) / M_B

This result in 2M_A/ M_B = (3/2) -1 = ½

OR M_B = 4M_A

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

4M_A = M_B