Question

At 300K, the density of a certain gaseous molecule at 2 bar is double that of dinitrogen (\[{{N}_{2}}\]) at 4 bar. The molar mass of gaseous molecule is:
(A) 112\[gmo{{l}^{-1}}\]
(B) 224\[gmo{{l}^{-1}}\]
(C) 28\[gmo{{l}^{-1}}\]
(D) 56 \[gmo{{l}^{-1}}\]

Answer 1

Hint: In this question we will consider the ideal condition and apply the ideal gas equation here. And also we know that the density of any gas depends on the pressure.

Step by step solution:
We know that ideal gas equation:
\[PV=nRT\] ……..(1)
And here, ‘P’ is the pressure of gas,
  ‘V’ is the volume of gas,
‘n’ is moles of gas
R is ideal gas constant
‘T’ is the temperature.
\[n=\dfrac{dV}{M}\] ……..(2)
Putting this value of ‘n’ in ideal gas equation (1):
\[P=\dfrac{dRT}{M}\] ……..(3)
And it is given that:
Temperature is constant, T= 300K
Pressure of unknown gas (\[{{P}_{unknown}}\]) = 2 bar
And pressure of nitrogen gas (\[{{P}_{nitrogen}}\]) = 4 bar
And also given that at these pressures and given temperature
Density of unknown gas = 2 x density of nitrogen gas
\[{{d}_{unknowngas}}\]= 2 x \[{{d}_{nitrogengas}}\]
If we apply ideal gas equation for both gases:
For unknown gas
\[{{P}_{unknown}}=\dfrac{{{d}_{unknown}}RT}{{{M}_{unknown}}}\] …….(4)
For nitrogen
\[{{P}_{nitrogen}}=\dfrac{{{d}_{nitrogen}}RT}{{{M}_{nitrogen}}}\] ……..(5)
Then, by dividing equation (4) by equation (5):
\[\dfrac{{{P}_{nitrogen}}}{{{P}_{unknown}}}=\dfrac{{{d}_{nitrogen}}\times {{M}_{unknown}}}{{{d}_{unknown}}\times {{M}_{nitrogen}}}\]
Now we will put given data:
\[\dfrac{4}{2}=\dfrac{{{d}_{nitrogen}}\times {{M}_{unknown}}}{2\times {{d}_{nitrogen}}\times 28}\]
\[{{M}_{unknown}}\]=112 \[gmo{{l}^{-1}}\]
So the correct answer is option “A”.

Note: Here you should remember that both known and unknown gases are ideal gases. In this question we need not to change the units of pressure because it will cancel out.