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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
For ideal gas A, the ideal gas equation is given by,
`"P"_"A""V" = "n"_"A""RT"` ............(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"_"s""V" = "n"_"s""RT"` ......(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"_"A""M"_"A")/"m"_"A" = ("RT")/"V"` .....(iii)
From equation (ii), we have
`"p"_"BV" = "m"_"B"/"M"_"B" "RT" => ("p"_"B""M"_"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"_"A""M"_"A")/"m"_"A" = ("p"_"B""M"_"B")/"m"_"B"` .....(v)
Given
`"m"_"A" = 1 "g"`
`"p"_"A" = 2 "bar"`
`"m"_"B" = 2"g"`
`"p"_"B" = (3 - 2) = 1 "bar"`
(Since total pressure is 3 bar)
Substituting these values in equation (v), we have
`(2xx"M"_"A")/1 = (1xx"M"_"B")/2`
`=> 4"M"_"A" = "M"_"B"`
Thus, a relationship between the molecular masses of A and B is given by
`4"M"_"A" = "M"_"B"`
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