Question

The inversion temperature ${T_i}$(K) of hydrogen is:
 [Given: Van der Waals constants ‘a’ and ‘b’ is \[0.244{\text{ }}atm\]${L^2}mo{l^{ - 2}}$ and \[0.027\]$Lmo{l^{ - 1}}$ respectively]
A) \[440\]
B) \[220\]
C) \[110\]
D) $330$

Answer 1

Hint: Inversion temperature is characteristic of each gas and is denoted by ${T_i}$ which basically is that temperature above which any gas will heats up upon expansion as the temperature increases and below this temperature any gas will cool down as the temperature decreases.


Complete solution:
We are well aware with the Inversion temperature which is that critical temperature which is involved in thermodynamics and cryogenics techniques above which a non-ideal gas expands at constant enthalpy and experiences an increase in temperature resulting in heating up of the gas and below which a temperature decrease resulting in cooling of the gas.
Also, Gases become cooler only if they are below a certain temperature in Joule Thomson’s expansion known as Inversion temperature (${T_i}$).
Now, we are given that:
\[a = 0.244{\text{ }}atm\,{L^2}mo{l^{ - 2}}\]
$b = 0.027Lmo{l^{ - 1}}$$R = 0.0821\,L\,atm\,{K^{ - 1}}mo{l^{ - 1}}$
After putting these values in the given formula we will get:
$\therefore {T_i} = \dfrac{{2\, \times \,0.244}}{{0.027 \times 0.0821}} = 220K$
Additional information:
The temperature change that is seen is known as the Joule-Thomson effect.
If we are taking van der Waals gas we can calculate the enthalpy $H$ by using statistical mechanics as
$H = \,\dfrac{5}{2}N{k_B}T + \dfrac{{{N^2}}}{V}(b{k_B}T - 2a)$
Where, $N$= number of molecules
  $V$= volume
  $T$= temperature
  ${k_B}$= Boltzmann’s constant
  a and b = constants which depends on intermolecular forces and molecular volumes respectively.
For the given equation, keeping the enthalpy constant and increasing volume, temperature will change depending on the sign of $b{k_B}T - 2a$ so, the equation becomes
${T_i} = \dfrac{{2a}}{{bR}}$

Therefore the correct answer is (B).


Note: Always remember that the Inversion temperature is seen during liquefaction of gases and it usually depends upon the nature of gas. Also the Boyle’s temperature is different from the Inversion temperature where a real gas behaves as an ideal gas over a particular temperature.