#### Question

The 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.

#### Solution 1

For ideal gas A, the ideal gas equation is given by,

`P_AV = n_ART` ............(i)

Where, *p*_{A} and *n*_{A} represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

`p_sV = n_sRT` ......(ii)

Where, *p*_{B }and *n*_{B} represent the pressure and number of moles of gas B.

[*V* and *T* are constants for gases A and B]

From equation (i), we have

`p_AV = m_A/M_A RT => (p_AM_A)/m_A = (RT)/V` .....(iii)

From equation (ii), we have

`p_BV = m_B/M_B RT => (p_BM_B)/m_B = RT/V` .......(iv)

Where, M_{A }and M_{B} are the molecular masses of gases A and B respectively.

Now, from equations (iii) and (iv), we have

`(p_M_A)/m_A = (p_BM_B)/m_B` .....(v)

Given

`m_A = 1 g`

`p_A = 2 "bar"`

`m_B = 2g`

`p_B = (3 -2) = 1 "bar"`

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

`(2xxM_A)/1 = (1xxM_B)/2`

`=> 4M_A = M_B`

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

`4M_A = M_B`

#### Solution 2

Suppose molecular masses of A and B are M_{A} and M_{B} respectively. Then their number of moles will be

`n_A = 1/M_A` , `n_B= 2/M_B`

`P_A = "2bar", P_A + P_B = 3 bar, i.e P_B = 1 "bar"`

Applying the relation PV = nRT

`P_AV = n_ART`, `P_BV = n_BRT`

`:. P_A/P_B = n_A/n_B = (1/M_A)/(2/M_B) = M_B/(2M_A)`

or `M_B/M_A = 2xx P_A/P_B = 2 xx 2/1 =\ 4 or M_B = 4 M_A`