Question

When helium is allowed to expand into a vacuum, the heating effect is observed. The reason for this is: (Assume ${\text{He}}$ as a non-ideal gas)
A. ${\text{He}}$ is an inert gas
B. The inversion temperature of Helium is very high
C. The inversion temperature of Helium is very low
D. ${\text{He}}$ has the lowest boiling point

Answer 1

The joule- thomson effect is shown by non-ideal gases or real gases. For ideal gases the value of the coefficient is zero. The formula of the of the joule thomson coefficient is given by $\mu_{jt}$ = -$(\dfrac{dT}{dP})_{H}$

Complete step by step answer:
-First of all let us analyze all the given data in question which says helium is a non-ideal gas and when it is allowed to expand into a vacuum the heating effect is observed.
-Now let us discuss the inversion temperature of helium where the helium gas Joule–Thomson inversion temperatures at a constant pressure of one atmosphere is very low.
-Therefore, the helium gas gets heated when expanded in to vacuum at constant enthalpy and at typical room temperatures.
Hence, option C is the correct choice of answer.

Additional Information: The non-ideal gas can be explained as a gas molecule that has no attraction for one another because they are situated at very far distance at relatively low pressures. But when the high pressure is applied the force of attraction is also no longer insignificant as at high pressure all the molecules are close to each other. The noble gases don’t get participated in chemical reactions with other gases.

Note: The hydrogen and helium are the two gases which give very low Joule–Thomson inversion temperatures at one atmosphere pressure. The inversion temperature of a gas is the critical temperature point where going below it a non-ideal gas (in this case helium) that is expanding at constant enthalpy will decrease the temperature and going above that point will increase the temperature.