
Which of the following options will have a compressibility factor greater than $1$ ?
A. ${{\text{H}}_{\text{2}}}$ gas at critical condition.
B. ${\text{C}}{{\text{H}}_{\text{4}}}$ gas at room temperature and low pressure
C. ${\text{N}}{{\text{H}}_4}$ gas at its Boyle’s temperature and low pressure
D. $He$ gas at normal temperature and normal pressure
Answer
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Hint:The compressibility is directly proportional to the volume. The strength of intermolecular forces changes the volume of the gas compared to the volume of the ideal gas. Volume in turn affects the compressibility factor.
Formula used: ${\text{Z = }}\,\dfrac{{{\text{pV}}}}{{{\text{nRT}}}}$
Complete answer:
The compressibility factor defines the deviation of the real gases from the ideal behaviour.
The formula of the compressibility factor is as follows:
${\text{Z = }}\,\dfrac{{{\text{pV}}}}{{{\text{nRT}}}}$
Where,
${\text{Z}}$ is the compressibility factor
${\text{p}}$ is the pressure
V is the volume
${\text{n}}$ is the number of moles
R is the gas constant
T is the temperature
For the ideal gas product of pressure and volume is equal to the product of the number of moles, gas constant, and temperature. So,
${\text{pV = }}\,{\text{nRT}}$
So, ${\text{Z = }}\,1$
So, the value of the compressibility factor for an ideal gas is one.
Due to small size hydrogen and helium both show intermolecular repulsion. Due to repulsion the actual volume of the hydrogen and helium gas is greater than the volume of an ideal gas.
So, the value of the compressibility factor for hydrogen and helium is greater than one.
${\text{Z > }}1$
So, hydrogen and helium both have a compressibility factor greater than one at critical condition.
$He$ gas is at normal temperature and normal pressure, not at critical condition, so only option A is correct.
Therefore, option (A) ${{\text{H}}_{\text{2}}}$gas at critical condition is correct.
Note:
The relation between compressibility and volume of ideal and real gas is as follows: ${\text{Z = }}\,\dfrac{{{{\text{V}}_{{\text{real}}}}}}{{{{\text{V}}_{{\text{ideal}}}}}}$ . When real gas behaves as ideal gas ${{\text{V}}_{{\text{real}}}} = \,{{\text{V}}_{{\text{ideal}}}}$ so, ${\text{Z = }}\,1$. When the intermolecular attraction forces are strong then the actual volume of the gas will be less than the volume of an ideal gas. The gas will have a compressibility factor, less than one. When the intermolecular repulsive forces are strong then the actual volume of the gas will be more than the volume of an ideal gas. The gas will have a compressibility factor greater than one.
Formula used: ${\text{Z = }}\,\dfrac{{{\text{pV}}}}{{{\text{nRT}}}}$
Complete answer:
The compressibility factor defines the deviation of the real gases from the ideal behaviour.
The formula of the compressibility factor is as follows:
${\text{Z = }}\,\dfrac{{{\text{pV}}}}{{{\text{nRT}}}}$
Where,
${\text{Z}}$ is the compressibility factor
${\text{p}}$ is the pressure
V is the volume
${\text{n}}$ is the number of moles
R is the gas constant
T is the temperature
For the ideal gas product of pressure and volume is equal to the product of the number of moles, gas constant, and temperature. So,
${\text{pV = }}\,{\text{nRT}}$
So, ${\text{Z = }}\,1$
So, the value of the compressibility factor for an ideal gas is one.
Due to small size hydrogen and helium both show intermolecular repulsion. Due to repulsion the actual volume of the hydrogen and helium gas is greater than the volume of an ideal gas.
So, the value of the compressibility factor for hydrogen and helium is greater than one.
${\text{Z > }}1$
So, hydrogen and helium both have a compressibility factor greater than one at critical condition.
$He$ gas is at normal temperature and normal pressure, not at critical condition, so only option A is correct.
Therefore, option (A) ${{\text{H}}_{\text{2}}}$gas at critical condition is correct.
Note:
The relation between compressibility and volume of ideal and real gas is as follows: ${\text{Z = }}\,\dfrac{{{{\text{V}}_{{\text{real}}}}}}{{{{\text{V}}_{{\text{ideal}}}}}}$ . When real gas behaves as ideal gas ${{\text{V}}_{{\text{real}}}} = \,{{\text{V}}_{{\text{ideal}}}}$ so, ${\text{Z = }}\,1$. When the intermolecular attraction forces are strong then the actual volume of the gas will be less than the volume of an ideal gas. The gas will have a compressibility factor, less than one. When the intermolecular repulsive forces are strong then the actual volume of the gas will be more than the volume of an ideal gas. The gas will have a compressibility factor greater than one.
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