Question

Compressibility factor for H2 behaving as real gas:
A.$1$
B.$\left( {1 - \dfrac{a}{{{\text{RTV}}}}} \right)$
C.$\left( {1 + \dfrac{{{\text{Pb}}}}{{{\text{RT}}}}} \right)$
D.$\dfrac{{{\text{RTV}}}}{{\left( {1 - {\text{a}}} \right)}}$

Answer 1

An ideal gas is one which obey or follows ideal gas equation ${\text{PV}}\,{\text{ = }}\,{\text{nRT}}$ under all conditions of temperature (T) and pressure (P). Whereas a real gas behaves as ideal gas only under high temperature or low pressure, whereas at high pressure and low temperature the volume occupied by the gas molecules is not negligible as compared to the total volume, also the force of attraction between the molecules is not negligible. So that real gas is said to obey Van der Waals equation.
Formula used:
Here we already know that the real gases obey Van der Waals equation, it is given by
$\left( {{\text{P}}\,{\text{ + }}\,\dfrac{{\text{a}}}{{{{\text{V}}^2}}}} \right)\, \times \,\left( {{\text{V - b}}} \right)\, = \,{\text{RT}}$
${\text{P - Pressure}} \\\ {\text{V - volume of gas}} \\\ {\text{R - universal gas constant}} \\\ {\text{T - Temperature}} \\\ {\text{a - correction factor which describes the measure of attraction between gas molecules}} \\\ {\text{b - correction factor for the volume taken by the gas molecules}} \\\$

Complete answer:
Now we discuss about the compressibility factor (Z) corrects the deviation from ideal gas law , so the accuracy of ideal gas equation ${\text{PV}}\,{\text{ = }}\,{\text{nRT}}$ can be determined by comparing the actual volume of ${\text{1 mole}}$ of gas to the molar volume of an ideal gas at same temperature and pressure. This ratio is called the compressibility factor.
The ideal gas law corrected for non-ideal gas is
${\text{PV}}\,{\text{ = }}\,{\text{ZRT}} \\\ {\text{Z - compressibility factor}} \\\$
So here we get ${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{PV}}}}{{{\text{RT}}}}$.
Complete step by step solution: So here we have to find the compressibility factor of H2 gas behaving as a real gas,
At high pressure and low temperature conditions the real gases doesn’t behave as ideal gas , so it follows Van der Waals equation
$\left( {{\text{P}}\,{\text{ + }}\,\dfrac{{\text{a}}}{{{{\text{V}}^2}}}} \right)\, \times \,\left( {{\text{V - b}}} \right)\, = \,{\text{RT}}$
At high pressure ${\text{P}}\,{\text{ + }}\dfrac{{\text{a}}}{{{{\text{V}}^2}}}\,\, \approx \,\,{\text{P}}$
So the Van der Waals equation becomes ${\text{P }} \times \,\left( {{\text{V - b}}} \right)\,\, = \,{\text{RT}}$
By rearranging the equation we get
$\dfrac{{{\text{P}}\left( {{\text{V - b}}} \right)}}{{{\text{RT}}}}\,\, = \,\,1 \\\ \dfrac{{{\text{PV}}\,}}{{{\text{RT}}}}\, - \,\dfrac{{{\text{P}}\,{\text{b}}}}{{{\text{RT}}}}\, = \,1 \\\$
We know the equation of compressibility factor , ${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{PV}}}}{{{\text{RT}}}}$. So by putting the value of Z on above equation we get
$Z - \dfrac{{{\text{P}}\,{\text{b}}}}{{{\text{RT}}}}\, = \,1$
Again by rearranging the above equation we get
${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{P}}\,{\text{b}}}}{{{\text{RT}}}}\, + \,1$
So here the compressibility factor of H2 gas behaving as real gas is ${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{P}}\,{\text{b}}}}{{{\text{RT}}}}\, + \,1$

Hence, the correct answer is option C.

Note:
Gases is said behave ideally when two conditions are met and the conditions are
The volume of gas molecules is small compared to the total volume
The forces between the gas molecules are not significant, only elastic collisions between the molecules will consider.
At high pressure and low temperature these conditions not be followed hence real gas doesn’t behave as ideal gas under high pressure or low temperature conditions