Question

For a 500K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then, what is the value of $\dfrac{b}{{RT}}$?

Answer 1

The compressibility factor (Z) is a beneficial thermodynamic property for changing the ideal gas law in order to account for behaviour of real gases. It refers to a measure of how much the thermodynamic properties of a real gas vary from that expected from an ideal gas. It may also be estimated as the ratio of the actual volume of a real gas to that volume as predicted by the ideal gas at the similar temperature and pressure as the actual volume.

Complete step by step answer:
Compressibility factor (Z), which is generally defined as $Z = \dfrac{{PV}}{{RT}}$(wherein P is pressure, V is the molar volume of gas, Z is compressibility factor, R is the universal gas constant and T is temperature), is always unity for an ideal gas. Though in case of high-pressure region, the expression for the compressibility factor becomes Z = 1 + \dfrac{{Pb}}{{RT}}$$$\;{10^{ - 3}}at{m^{ - 1}}$$. Thus when Z=2 and P=1000 atm, the expression becomes as follows: 2 = 1 + \dfrac{{1000b}}{{RT}}$$\dfrac{b}{{RT}} = \dfrac{{2 - 1}}{{1000}} = {10^{ - 3}}at{m^{ - 1}} Thus, when Z = 2.2 and P = 1200 atm, the expression becomes: $$2.2 = 1 + \dfrac{{1200b}}{{RT}}$$ $$\dfrac{b}{{RT}} = \dfrac{{2.2 - 1}}{{1200}} = {10^{ - 3}}at{m^{ - 1}}$$ Hence, the value of\dfrac{b}{{RT}}$is ${10^{ - 3}}at{m^{ - 1}}$.

Note:
The compressibility factor should not be confused with the coefficient of isothermal compressibility. In most engineering works, the compressibility factor is generally employed as a correction factor to ideal behaviour. Therefore, ${v_{real}} = Z{v_{id}}$is employed to determine the actual volume, ${v_{real}}$by multiplying the compressibility factor with the ideal gas volume, at the same temperature and pressure.