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, pA and nA 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, pB and nB 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, MA and MB 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 MA and MB 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`
