
For an ideal gas, the Joule-Thompson coefficient is-
A. zero
B. positive
C. negative
D. depends on atomicity of gas
Answer
133.5k+ 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.
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.
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 gas | Sign 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.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Thermodynamics Class 11 Notes: CBSE Chapter 5
