Question

The compressibility factor of an ideal gas is
(A 4
(B) 8
(C) 1
(D) 2

Answer 1

. We know that for an ideal gas, the actual molar volume of the gas is equal to the calculated molar volume at the same temperature and pressure. Also, compressibility factor (Z) is defined as the ratio of molar volume of given gas to that of molar volume of that gas at ideal conditions.
$Z=\dfrac{{{V}_{r}}}{{{V}_{i}}}$
In the above formula, ${{V}_{r}}$ is the molar volume of given gas and ${{V}_{i}}$ is the molar volume of given gas in ideal condition.

Complete step by step answer:
-Compressibility factor is defined as the ratio of molar volume of gas to that of ideal gas. It is the measurement of the amount a gas is deviating from its ideal behavior at similar pressure and temperature. It is represented by the symbol 'Z'. There are no attractive or repulsive forces between the molecules of an ideal gas. So, it ${{V}_{i}}$ will be equal to ${{V}_{r}}$.
-Therefore, on applying the above formula, we will get a compressibility factor of above gas that is equal to 1.
So, the correct answer is “Option C”.

Additional Information:
-In real gas, it is seen that pressure and molecular attraction tend to confine the molecules while temperature and molecular repulsions tend to separate them.
-Whenever the temperature of gas is high, then kinetic energy increases and molecules tend to move apart. So, at high temperatures, the actual volume of gas will be more than the ideal gas.

Note: -It has been seen that those gases which have gas compressibility factor is equal to1 behaves in a similar manner like an ideal gas.
-At high temperature, $Z>1$ for real gas and repulsive forces dominates due to which gas cannot be liquefied.
-At low pressure, $Z<1$ for real gases and attractive forces dominate due to which gas can be easily liquefied.
-At standard conditions, $Z<1$ for all real gases except hydrogen and helium gas.