Question

Match column 1 with column 2 and select the correct answer.

Column-1	Column-2
A) Critical temperature	1) $\dfrac{a}{{Rb}}$
B) Boyle's temperature	2) $\dfrac{{2a}}{{Rb}}$
C) Inversion temperature	3) $\dfrac{T}{{{T_C}}}$
D) Reduced temperature	4) $\dfrac{{8a}}{{27Rb}}$

A. A-2, B-1, C-4 ,D-3
B. A-4, B-3, C-2, D-1
C. A-2, B-3, C-4 , D-1
D. A-4, B-1, C-2, D-3

Answer 1

Study about the critical temperature, Boyle's temperature, Inversion temperature and Reduced temperature of gas and the expression of them to match the column

Formula used:
Critical temperature of van der Waals gas is given by,
${T_C} = \dfrac{{8a}}{{27Rb}}$
where, $a$ and $b$ are the constant of van der Waals equation and $R$ is the universal molar gas constant.
The virial equation of state for real gas is given by,
$P = RT\left( {\dfrac{1}{{{V_m}}} + \dfrac{{{B_2}(T)}}{{V_m^2}} + \ldots } \right)$
where, $P$ is the pressure, $R$ is the universal molar gas constant, $T$ is the temperature of the gas, ${V_m}$ is the volume of the gas and ${B_3}(T),{B_3}(T),{B_3},(T)....$ are the virial coefficients.
Boyle's temperature is given by, ${T_b} = \dfrac{a}{{Rb}}$.
Inversion temperature is given by, ${T_i} = \dfrac{{2a}}{{Rb}}$
Reduced temperature is given by, ${T_r} = \dfrac{T}{{{T_C}}}$
where $T$ is the absolute temperature of the gas.

Complete step by step answer:
We know that the critical temperature of any substance is the maximum temperature at which the substance can be liquefied, above the critical temperature the substance cannot be liquefied at any pressure. Critical temperature for van der Waals gas is given by, ${T_C} = \dfrac{{8a}}{{27Rb}}$. Where, $a$ and $b$ are the constant of van der Waals equation and $R$ is the universal molar gas constant.

We know for a real gas equation of state can be written by the virial equation, $P = RT\left( {\dfrac{1}{{{V_m}}} + \dfrac{{{B_2}(T)}}{{V_m^2}} + \ldots } \right)$ where, $P$ is the pressure, $R$ is the universal molar gas constant, $T$ is the temperature of the gas, ${V_m}$ is the volume of the gas and ${B_3}(T),{B_3}(T),{B_3},(T)....$ are the virial coefficients. Boyle's temperature is the temperature at which the second coefficient of virial equation ${B_2}(T)$ becomes zero.

For this condition we get Boyle's temperature as, ${T_b} = \dfrac{a}{{Rb}}$. We know Inversion temperature is the temperature for real gas below which expanding gas experiences a decrease in temperature and above which it experiences an increase in temperature at constant enthalpy. Inversion temperature is given by, ${T_i} = \dfrac{{2a}}{{Rb}}$. Also, we know reduced temperature is the ratio of the absolute temperature and the critical temperature and is given by, ${T_r} = \dfrac{T}{{{T_C}}}$ where $T$ is the absolute temperature of the gas. So, we can match the columns as,

Column-1	Column-2
A) Critical temperature	4)$\dfrac{{8a}}{{27Rb}}$
B) Boyle's temperature	1)$\dfrac{a}{{Rb}}$
C) Inversion temperature	2)$\dfrac{{2a}}{{Rb}}$
D) Reduced temperature	3)$\dfrac{T}{{{T_C}}}$

Hence,the correct answer is option D.

Note: The different thermodynamic temperatures for van der Waals gas and their expression must be memorized to solve the problem easily. Since, it is too complex to solve for the expression for each thermodynamic temperature, because it contains rigorous mathematical calculations.