Home › Physics › Thermal Physics › Joule-Thomson Effect
Thermal Physics
Joule-Thomson Effect
The Joule-Thomson effect describes the temperature change of a real gas when it undergoes steady, adiabatic expansion from a high pressure region to a low pressure region through a porous plug or throttling valve. During this process, the total enthalpy of the system remains constant.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of Joule-Thomson effect, inversion temperature, qualitative behavior of real and ideal gases, and applications such as liquefaction of gases.
PGT extension
Mathematical derivation of the Joule-Thomson coefficient using Maxwell's relations, expression for van der Waals gases, inversion temperature derivation, and proof that the coefficient is zero for ideal gases.
Prerequisites
First and Second Laws of Thermodynamics
Concept of Enthalpy and Adiabatic Processes
Van der Waals Equation of State for Real Gases
Fundamentals of the Joule-Thomson Effect
It is a throttling process characterized by constant enthalpy (isenthalpic process), meaning H1 = H2.
When a real gas expands adiabatically from high pressure to low pressure through a porous plug, it may undergo either cooling or heating depending on the initial temperature.
For an ideal gas, the Joule-Thomson coefficient is identically zero because internal energy depends only on temperature and there are no intermolecular forces.
Joule-Thomson Coefficient
Defined as the rate of change of temperature with respect to pressure at constant enthalpy: μJT = (∂T / ∂P)H.
If μJT > 0, the gas cools upon expansion (cooling region).
If μJT < 0, the gas heats up upon expansion (heating region).
Inversion Temperature
The temperature at which the Joule-Thomson coefficient changes sign is known as the inversion temperature (Ti).
At the inversion temperature, μJT = 0, and the gas neither cools nor warms upon throttling.
A gas shows cooling only when its temperature is below its inversion temperature for a given pressure.
Behavior of Real Gases (Van der Waals Gases)
At temperatures below the Boyle temperature, real gases show cooling at low pressures and heating at high pressures.
The maximum inversion temperature for a van der Waals gas is given by Ti,max = 2a / (bR), where a and b are van der Waals constants and R is the universal gas constant.
Hydrogen and helium have very low inversion temperatures (below room temperature), hence they must be pre-cooled below their inversion temperatures before they can be liquefied by throttling.
Derivations
Derivation of Joule-Thomson Coefficient using Maxwell Relations
Conditions: Single-component homogeneous system undergoing a constant enthalpy (throttling) process.
Start with the fundamental thermodynamic property relation for enthalpy: d$H = TdS + VdP$.
Express entropy S as a function of temperature T and pressure P, so that d$S = (\partial S / \partial T)_{P} dT + (\partial S / \partial P)_{T} dP$.
Substitute dS into the enthalpy equation: d$H = T [(\partial S / \partial T)_{P} dT + (\partial S / \partial P)_{T} dP] + V dP$.
Use the Maxwell relation (∂S / ∂P)T = -(∂V / ∂T)P and the heat capacity at constant pressure CP = T(∂S / ∂T)P.
Rewrite the enthalpy differential: d$H = C_{P} dT + [V - T(\partial V / \partial T)_{P}] dP$.
For an isenthalpic process (dH = 0), set the equation to zero and solve for dT / dP at constant H.
Conditions: Low pressures and moderate temperatures where the van der Waals equation applies.
Write the van der Waals equation of state: (P + \frac{a}{V^2})(V - b) = RT.
Expand the equation neglecting higher-order terms in b: PV + \frac{a}{V} - Pb - \frac{ab}{$V^2 = RT \implies V \approx \frac{RT}{P} + b - \frac{a}{RT}$.
Differentiate volume with respect to temperature at constant pressure: \left(\frac{\partial V}{\partial T}\right)_{P} = \frac{R}{P} + \frac{a}{RT^2}.
Substitute this derivative into the general thermodynamic expression for \mu_{JT}: \m$u_{JT} = \frac{1}{C_P} \left[ T \left( \frac{R}{P} + \frac{a}{RT^2} \right) - V \right]$.
Substitute V \approx \frac{RT}{P} + b - \frac{a}{RT} into the expression to simplify terms.
Obtain the final expression for the van der Waals Joule-Thomson coefficient.
Result: \m$u_{JT} = \frac{1}{C_{P}} \left( \frac{2a}{RT} - b \right)$
Draw: Curves of constant enthalpy plotting Pressure (P) on the y-axis against Temperature (T) on the x-axis, showing a parabola-like envelope enclosing the inversion curve.
Interpret: The peak of each isenthalpic curve represents the inversion temperature at that pressure. Inside the inversion parabola, μJT > 0 (cooling); outside, μJT < 0 (heating).
Exam use: Frequently asked in PGT conceptual multiple-choice questions regarding regions of heating and cooling during expansion.
porous plug experiment (Regnault / Joule-Kelvin)
Draw: A thermal insulated tube with a porous plug in the middle, fitted with upstream and downstream manometers and thermometers.
Interpret: Measures the steady-state temperature difference (ΔT) across the plug for a known pressure drop (ΔP) under adiabatic conditions.
Exam use: Direct questions on experimental setup, thermal insulation requirements, and verification of isenthalpic nature.
Joule-Thomson Coefficient vs Temperature
Draw: A graph of μJT on the y-axis versus Temperature T on the x-axis, intersecting the zero axis at Ti.
Interpret: Shows positive values at low temperatures (cooling), crosses zero at $T = T_{i}$, and becomes negative at high temperatures (heating).
Exam use: Testing understanding of inversion temperature and sign conventions of thermal coefficients.
Solved numericals
TGT core
Why does an ideal gas show no temperature change during a Joule-Thomson expansion?
Recall the expression for the Joule-Thomson coefficient: \m$u_{JT} = \frac{1}{C_{P}} \left[ T\left(\frac{\partial V}{\partial T}\right)_{P} - V \right]$.
For an ideal gas obeying PV = RT, find \left(\frac{\partial V}{\partial T}\right)_{P} = \frac{R}{P} = \frac{V}{T}.
Substitute this into the bracket term: T(V / T) - $V = V - V = 0$.
Therefore, \m$u_{JT} = 0$, indicating no temperature change (ΔT = 0).
Answer: Delta T = 0 because internal energy of an ideal gas depends solely on temperature and there are no intermolecular forces.
PGT extension
Calculate the inversion temperature for a gas obeying the van der Waals equation, given that the van der Waals constants are $a = 0.364 N m^4 / mol^2$ and $b = 4.27 \times 10^{-5} m^3 / mol$, and $R = 8.314 J / mol K$.
State the formula for maximum inversion temperature: $T_{i} = \frac{2a}{bR}$.
Substitute the given values: a = 0.364, $b = 4.27 \times 10^{-5}$, R = 8.314.
A real gas has a van der Waals constant $a = 0.140 N m^4 / mol^2$ and $b = 3.91 \times 10^{-5} m^3 / mol$. Determine whether the gas will cool or heat when throttled from high pressure to low pressure at room temperature (300 K). Assume $C_P = 20.8 J / mol K$.
Calculate the Joule-Thomson coefficient using the formula: \m$u_{JT} = \frac{1}{C_{P}} \left( \frac{2a}{RT} - b \right)$.
Substitute the values: $R = 8.314 J/mol K$, T = 300 K, $C_P = 20.8 J/mol K$.
Compute the first term inside the parenthesis: \frac{2 \times 0.140}{8.314 \times 300} = \frac{0.280}{2494.2} \approx 1.122 \times 10^{-4} m^3/mol.
Since \mu_{JT} > 0, the temperature drops (gas cools).
Answer: m$u_{JT} = +3.51 \times 10^{-6} K/Pa (The gas cools upon throttling)$.
Assumptions and limitations
The process is strictly adiabatic; no heat enters or leaves the system through the walls or the porous plug.
Steady-state flow conditions are maintained across the throttling device.
Kinetic energy changes of the gas before and after the plug are assumed to be negligible.
Van der Waals approximation fails at very high pressures where higher-order virial coefficients become significant.
Common misconceptions
Common error: Joule-Thomson expansion is an adiabatic free expansion.
Correct: It is a steady-flow throttling process through a porous plug maintaining constant enthalpy, whereas free expansion is characterized by constant internal energy (U = constant).
In throttling, external work is done by the upstream fluid and on the downstream fluid, keeping enthalpy constant rather than internal energy.
Common error: All gases cool down when expanded through a valve.
Correct: Gases heat up if their temperature is above their inversion temperature.
The sign of the Joule-Thomson coefficient depends on temperature relative to Ti; above Ti, μJT is negative.
Common error: Hydrogen and helium cool at room temperature during throttling.
Correct: Hydrogen and helium heat up at room temperature.
Their inversion temperatures are very low (around 204 K for H2 and 43 K for He), so at 300 K they are well above Ti.
Quick revision
Joule-Thomson effect is an isenthalpic process (H = constant).
Ideal gases show no Joule-Thomson cooling or heating (μJT = 0).
Inversion temperature is the temperature where μJT changes sign.
Cooling occurs when T < Ti (μJT > 0); heating occurs when T > Ti (μJT < 0).
Maximum inversion temperature for van der Waals gas is Ti = 2a / (bR).
Must remember
The Joule-Thomson process is an isenthalpic process characterized by constant enthalpy ($H_1 = H_2$).
The Joule-Thomson coefficient is defined as $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$.
For an ideal gas, the Joule-Thomson coefficient $\mu_{JT} = 0$ because there are no intermolecular forces.
The maximum inversion temperature for a van der Waals gas is given by $T_{i} = \frac{2a}{bR}$.
Hydrogen and helium show a heating effect at room temperature because their inversion temperatures are well below room temperature.
Glossary
Joule-Thomson Coefficient ($\mu_{JT}$)The rate of change of temperature with respect to pressure at constant enthalpy during a throttling process.
Inversion Temperature ($T_i$)The temperature at which the Joule-Thomson coefficient changes sign and equals zero.
Isenthalpic ProcessA thermodynamic process that proceeds at constant enthalpy ($H$).
Boyle Temperature ($T_B$)The temperature at which a real gas behaves like an ideal gas over a range of pressures, given by $T_B = \frac{a}{bR}$.
Comparison of Expansion Types
Property
Joule-Thomson Expansion
Adiabatic Free Expansion
Constant Quantity
Enthalpy ($H = \text{constant}$)
Internal Energy ($U = \text{constant}$)
Work Done
Non-zero flow work
Zero work done ($W = 0$)
Ideal Gas Temperature Change
Zero temperature change
Zero temperature change
Behavior of Gases based on $\mu_{JT}$
Condition
Sign of $\mu_{JT}$
Effect on Expansion
$T < T_i$
Positive ($\mu_{JT} > 0$)
Cooling effect
$T > T_i$
Negative ($\mu_{JT} < 0$)
Heating effect
$T = T_i$
Zero ($\mu_{JT} = 0$)
Neither cooling nor heating
Practice Quiz
Choose an option to reveal the explanation.
1. Which thermodynamic property remains constant during a Joule-Thomson throttling process?
Explanation: The Joule-Thomson process is an adiabatic throttling process characterized by constant enthalpy ($H_1 = H_2$).
2. What is the value of the Joule-Thomson coefficient $\mu_{JT}$ for an ideal gas?
Explanation: For an ideal gas, internal energy depends solely on temperature and intermolecular forces are absent, making $\mu_{JT} = 0$.
3. How is the Joule-Thomson coefficient mathematically defined?
Explanation: By definition, the Joule-Thomson coefficient is $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$, representing the rate of temperature change with pressure at constant enthalpy.
4. If the Joule-Thomson coefficient $\mu_{JT} > 0$ for a gas, what happens to the gas upon adiabatic expansion through a porous plug?
Explanation: When $\mu_{JT} > 0$, $\Delta T$ is negative for a pressure drop ($\Delta P < 0$), resulting in cooling.
5. What is the maximum inversion temperature $T_i$ for a van der Waals gas in terms of constants $a$, $b$, and $R$?
Explanation: The maximum inversion temperature for a van der Waals gas is derived as $T_i = \frac{2a}{bR}$.
6. Why do hydrogen and helium heat up when expanded through a porous plug at room temperature (300 K)?
Explanation: Hydrogen and helium possess very low inversion temperatures (around 204 K and 43 K respectively). At 300 K, they are above their inversion temperatures, yielding $\mu_{JT} < 0$ (heating effect).
7. The thermodynamic identity expressing the Joule-Thomson coefficient in terms of volume, temperature, and molar heat capacity $C_P$ is:
Explanation: Using Maxwell's relations, the general expression for any pure substance is $\mu_{JT} = \frac{1}{C_P}\left[T\left(\frac{\partial V}{\partial T}\right)_P - V\right]$.
8. At the inversion temperature, the value of the Joule-Thomson coefficient $\mu_{JT}$ is:
Explanation: The inversion temperature is defined as the temperature at which $\mu_{JT}$ changes sign and is precisely equal to zero.
9. What is the relation between the Boyle temperature ($T_B$) and the maximum inversion temperature ($T_i$) for a van der Waals gas?
Explanation: Since $T_B = \frac{a}{bR}$ and $T_i = \frac{2a}{bR}$, it follows directly that $T_i = 2 T_B$.
10. Which of the following processes describes adiabatic free expansion?
Explanation: In an adiabatic free expansion into a vacuum, no work is done and no heat enters or leaves, meaning internal energy remains constant ($U = \text{constant}$).
11. For a van der Waals gas, the expression for the Joule-Thomson coefficient is given by:
Explanation: Substituting the van der Waals equation of state into the general thermodynamic identity yields $\mu_{JT} = \frac{1}{C_P}\left(\frac{2a}{RT} - b\right)$.
12. What must be done to liquefy hydrogen or helium using the Joule-Thomson effect?
Explanation: Because their inversion temperatures are below room temperature, H₂ and He must be pre-cooled below $T_i$ before throttling produces cooling and eventual liquefaction.
13. In the cooling region of the Joule-Thomson effect, as pressure decreases across a porous plug ($\Delta P < 0$), what happens to the temperature?
Explanation: Since $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H > 0$, a decrease in pressure ($\Delta P < 0$) leads to a decrease in temperature ($\Delta T < 0$).
14. The units of the Joule-Thomson coefficient $\mu_{JT}$ are:
Explanation: Since $\mu_{JT} = \frac{\partial T}{\partial P}$, its units are temperature over pressure, i.e., Kelvin per Pascal or Kelvin per atmosphere.
15. What condition characterizes the boundary of the inversion curve where a gas transitions from cooling to heating?
Explanation: The inversion curve represents the locus of states where $\mu_{JT} = 0$, separating the cooling region from the heating region.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The Joule-Thomson coefficient for an ideal gas is:
Explanation: An ideal gas has no intermolecular forces, hence its temperature does not change during throttling, making $\mu_{JT} = 0$.
2. Which of the following gases warms up when allowed to expand adiabatically through a porous plug at room temperature?
Explanation: Hydrogen has an inversion temperature around 204 K. At room temperature (300 K), it is above $T_i$, so it heats up upon expansion.
3. The maximum inversion temperature for a van der Waals gas is given by:
Explanation: Setting $\mu_{JT} = \frac{1}{C_P}\left(\frac{2a}{RT} - b\right) = 0$ gives the maximum inversion temperature $T_i = \frac{2a}{bR}$.
4. During a Joule-Thomson throttling process, which of the following remains constant?
Explanation: Throttling is an isenthalpic process, meaning enthalpy remains constant ($H_1 = H_2$).
5. The Joule-Thomson coefficient $\mu_{JT}$ is defined by the partial derivative:
Explanation: By definition, $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$.
6. If a gas expands under conditions where $\mu_{JT} < 0$, the gas will:
Explanation: When $\mu_{JT} < 0$, a pressure drop ($\Delta P < 0$) results in a temperature rise ($\Delta T > 0$), meaning the gas heats up.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. In a throttling process, fluid flows steadily through a porous plug. What is the net work done on or by the fluid across the device?
Explanation: Flow work is done by the upstream fluid ($P_1V_1$) and on the downstream fluid ($P_2V_2$), which combines with internal energy to keep enthalpy constant.
2. For a real gas obeying $\mu_{JT} = \frac{1}{C_P}\left(\frac{2a}{RT} - b\right)$, at what temperature does the cooling effect transition to heating?
Explanation: Setting $\mu_{JT} = 0$ gives $\frac{2a}{RT} - b = 0$, which yields $T = \frac{2a}{bR}$ (the inversion temperature).
3. Which of the following statements is correct regarding the inversion temperature of real gases?
Explanation: The inversion temperature depends on the applied pressure, forming an inversion curve that separates the cooling and heating regions.
4. What is the primary reason why hydrogen gas warms up upon throttling at room temperature?
Explanation: Because room temperature (300 K) exceeds hydrogen's inversion temperature (204 K), $\mu_{JT}$ is negative, causing heating.
5. In terms of the coefficient of volume expansion $\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$, the Joule-Thomson coefficient can be written as:
Explanation: Since $\left(\frac{\partial V}{\partial T}\right)_P = V \alpha$, substituting this into the general identity gives $\mu_{JT} = \frac{V}{C_P}(\alpha T - 1)$.
6. How does adiabatic free expansion differ fundamentally from Joule-Thomson throttling?
Explanation: Adiabatic free expansion is characterized by constant internal energy ($U = \text{constant}$), whereas throttling is an isenthalpic process ($H = \text{constant}$).
7. At temperatures well above the inversion temperature, a real gas will:
Explanation: When $T > T_i$, the Joule-Thomson coefficient $\mu_{JT}$ is negative, meaning expansion causes the temperature to increase (heating).
8. If a gas has a Boyle temperature $T_B$, what is its maximum inversion temperature approximately equal to?
Explanation: Since $T_B = \frac{a}{bR}$ and $T_i = \frac{2a}{bR}$, the maximum inversion temperature is twice the Boyle temperature ($2 T_B$).
9. Why is it impossible to liquefy helium by simple throttling at standard ambient temperature?
Explanation: Helium's inversion temperature is approximately 43 K. At ambient temperatures, $\mu_{JT}$ is negative (it heats up), so pre-cooling below 43 K is mandatory.
10. Which Maxwell relation is most directly useful in deriving the general thermodynamic identity for $\mu_{JT}$?
Explanation: Maxwell's relation $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$ is employed when transforming the expression for enthalpy variations during expansion.