Question

Explain the term compressibility factor ‘Z’. Solve the given problem; 20 ml of chlorine gas occupies a volume of 800ml at 300K and $5 \times {10^6}{\text{Pa}}$pressure. Calculate the compressibility factor of the gas (R=0.083L bar K-1 mol-1). State whether gas is more compressible or less.

Answer 1

compressibility factor (Z) is the measure of deviation of the real gases from the ideal gas behaviour and it can be defined as the ratio of product of pressure (p) and volume (V) to the product of number of moles (n), universal gas constant(R) and temperature (T). Mathematically it can be written as
$Z = \dfrac{{pV}}{{nRT}}$
For ideal gases Z=1 at temperature and pressure. For real gases it can be greater than one or less than one for the given temperature or pressure. If Z>1, it is very difficult to compress the gas and if Z<1, the gas is liquefiable.

Complete step by step answer:

1. Using the above equation the compressibility factor (Z) can be calculated, but if we observe the given universal gas constant(R) value the units of which is not matching with the units of given values of pressure, volume and number of moles, so let’s convert the units and then substitute the values in the equation.
2. Pressure is given in Pascal, it needs to be converted into a bar. $5 \times {10^6}Pa = 50bar$
3. Volume is given in ml; it should be converted to Liter. $800ml = 800 \times {10^{ - 3}}L$
4. Amount of chlorine is given in terms of ml; it should be converted to moles. $20ml = 0.88 \times {10^{ - 3}}mol$
5. Now substitute all the values in the above equation
   $Z = \dfrac{{50bar \times 800 \times {{10}^{ - 3}}L}}{{0.881 \times {{10}^{ - 3}}mol \times 0.083Lbar{K^{ - 1}}mo{l^{ - 1}} \times 300K}} \\\ Z = 1823.4116 \\\$
6. ‘Z’ value is greater than one therefore it is very difficult to compress the gas at given temperature and pressure.
7. Number of moles: we know that 1mole of any gas occupies 22.7L of volume. Mathematically written as
   ${\text{ }}22.7L = 1mole{\text{ }}of{\text{ }}C{l_2} \\\ 22.7 \times 1000ml = 1mole{\text{ }}of{\text{ }}C{l_2} \\\ {\text{ }}22.7 \times {10^3}ml = 1mole{\text{ }}of{\text{ }}C{l_2} \\\ {\text{ }}20ml = \dfrac{{20ml \times 1mole}}{{22.7 \times {{10}^3}ml}} = 0.881 \times {10^{ - 3}}mole \\\$
   Therefore 20ml of chlorine gas contains $0.881 \times {10^{ - 3}}mole{\text{ }}$of chlorine molecules.

Note: Conversions of units:

1. Pressure:
   $1 \times {10^5}{\text{Pa = 1bar}} \\\ 5 \times {10^6}Pa = \dfrac{{{\text{1bar}} \times 5 \times {{10}^6}Pa}}{{1 \times {{10}^5}{\text{Pa}}}} = 50bar \\\$
2. Volume:
   $1000ml = 1L \\\ 800ml = \dfrac{{1L \times 800ml}}{{1000ml}} = 800 \times {10^{ - 3}}L \\\$