Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

For an ideal gas, the Joule-Thompson coefficient is-
A. zero
B. positive
C. negative
D. depends on atomicity of gas

seo-qna
Last updated date: 19th Jun 2024
Total views: 53.4k
Views today: 0.53k
Answer
VerifiedVerified
53.4k+ views
Hint: The Joule-Thomson coefficient is related to Joule-Thomson effect, which is also known as Kelvin–Joule effect .It is further related to the change in temperature of fluid while flowing from a higher pressure region to a region of lower pressure.

Step-by-step solution:
The Joule-Thomson effect talks about the change in temperature that a liquid undergoes, when it is made to flow with force through a valve. It is made sure that the entire system is in proper insulation during this process, so that there is no exchange of heat with the environment.
This effect is only valid for real gases or liquids.
Joule-Thomson coefficient is defined as the rate of change of temperature T with respect to pressure P, at constant enthalpy H. it is represented as$\mu_{JT}$.
This coefficient can be expressed in terms of the volume of gas V, its coefficient of thermal expansion ∝ and its heat capacity at constant pressure $C_P$
Numerically, Joule-Thomson coefficient can be written as, $\mu_{JT} = (\partial T/\partial P)_H = (V/C_P)(\alpha T - 1)$
The value of $\mu_{JT}$ depends upon- nature of gas, temperature of gas before expansion, pressure of gas before expansion.
All real gases possess an inversion point at which the value of $\mu_{JT}$ changes its sign.

Temperature of gasSign of $\mu_{JT}$ $\partial P$ $\partial T$ Effect on gas
Below inversion temperature positive Always negative negative Cooling of gas
Above inversion temperature negative Always negative positive Warming of gas


In case of ideal gases, $\mu_{JT}$ is always zero, because they neither warm nor cool upon expansion at constant enthalpy. This implies that $\partial T = 0$ and thus .$\mu_{JT} = 0$. .

So, the correct option is A.

Note: The cooling effect produced in the Joule–Thomson expansion effect makes it a valuable tool in refrigeration. In the petrochemical industries, this effect is used as a standard process to liquefy gases. This method also finds its application in the production of liquid oxygen, nitrogen.