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.

Explanation

Suppose the molecular mass of the ideal gas A is MA and that of B is Me. When only the ideal gas A is present,
`PV = nRT`
`2xx V = 1/M_Axx RxxT (:. n = m/M =1/M) ...(i)`
On mixing the two gases, total number of moles
`=1/M_A+2/M_B`
Hence, according to the gas equation
`PV = nRT`
`3xxV=(1/M_A+2/M_B)xxRxxT ... (ii)`
Dividing eq. (ii) by eq. (i), we have
`3/2 = ((M_B + 2M_A)/(M_A M_B))/(1/M_A)=(M_B+2M_A)/M_B =(1+2xxM_A)/M_B`
or `2xxM_A/M_B = 3/2 - 1 = 1/2`
or `M_A/M_B = 1/2 xx 1/2 = 1/4`
or `M_B =4xxM_A`