Question

Calculate the compressibility factor for $C{{O}_{2}}$, if one mole of it occupies 0.4 litres at 300K and 40atm. Comment on the result.
(A) 0.40, $C{{O}_{2}}$ is more compressible than the ideal gas
(B) 0.65,$C{{O}_{2}}$ is more compressible than the ideal gas
(C) 0.55,$C{{O}_{2}}$ is more compressible than the ideal gas
(D) 0.62,$C{{O}_{2}}$ is more compressible than the ideal gas

Answer 1

Hint: To solve this you can use the formula of compressibility factor which is, \[Z=\dfrac{PV}{nRT}\].
If the calculated compressibility factor is less than one then the gas is more compressible than ideal gas and if it is more than one, then the gas is less compressible than ideal gas.

Complete step by step solution:
To answer this question, firstly we have to understand what the compressibility factor is.
We also know compressibility factor as the gas deviation factor and we denote it a Z. It is basically a correction factor which is introduced to describe the deviation of a real gas from an ideal gas behaviour at the same conditions of temperature and pressure.
The compressibility factor is simply given by the ratio of the molar volume of the gas to the molar volume of the ideal gas at the same constant temperature and pressure.
For an ideal gas, the compressibility factor is 1. If the compressibility factor of gas is lower than 1 i.e. that of the ideal gas then the gas is considered to be more compressible than ideal gas. Similarly, if the compressibility factor of gas is higher than 1, then the gas is considered to be less compressible than ideal gas.
We can calculate the compressibility factor in the given conditions for carbon dioxide gas using the formula-
\[Z=\dfrac{PV}{nRT}\]
where, Z is the compressibility factor, P is the pressure, T is the temperature, R is the universal gas constant, T is the temperature, V is the volume.
Now, let us calculate the compressibility factor for carbon dioxide.
n = 1 mole, T = 300K, V = 0.4L, P = 40atm and R = 0.082 Latm/molK
Therefore, \[Z=\dfrac{40atm\times 0.4L}{1mol\times 0.082Latm/molK\times 300K}=0.65\]
We can see from the above calculation that the compressibility factor for carbon dioxide is 0.65. It is less than 1 therefore it is more compressible than ideal gas.

Therefore, the correct answer is option [B] 0.65,$C{{O}_{2}}$ is more compressible that ideal gas.

Note: We should not confuse compressibility factor with compressibility which is the measure of the change in volume of a fluid or solid material when introduced to a change in pressure. This thermodynamic property is of great use as it helps us to modify the ideal gas law to account for real gas behaviour. The deviation of a real gas from an ideal gas is significant when the gas in on the edge of phase change i.e. low temperature or/and high pressure.