Home › Physics › Thermal Physics › Porous Plug Experiment and Joule-Thomson Effect

Thermal Physics

Porous Plug Experiment and Joule-Thomson Effect

The porous plug experiment demonstrates the Joule-Thomson effect, wherein a gas undergoes a temperature change when forced adiabatically through a porous plug from a region of high pressure to low pressure. This phenomenon reveals the departure of real gases from ideal gas behavior due to intermolecular forces.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Qualitative understanding of the Joule-Thomson effect, cooling of real gases, inversion temperature, and conditions for cooling versus heating.

PGT extension

Thermodynamic treatment proving constant enthalpy (isenthalpic process), Joule-Thomson coefficient derivation for real gases (Van der Waals equation), and inversion temperature relations.

Prerequisites

First Law of Thermodynamics for flow processes
Concept of enthalpy and adiabatic expansion
Van der Waals equation of state for real gases

Principle of the Porous Plug Experiment

A gas is forced at a constant high pressure through a porous plug into a region of lower pressure inside an adiabatically walled tube.
The process is steady-flow and adiabatic, ensuring no heat exchange with the surroundings.
Work is done by the gas on the low-pressure side and on the gas by the high-pressure side, resulting in an isenthalpic process.

Joule-Thomson Coefficient

Defined as the rate of change of temperature with pressure at constant enthalpy: \m$u_{JT} = (\frac{\partial T}{\partial P})_H$.
A positive Joule-Thomson coefficient (\mu_{JT} > 0) indicates cooling upon expansion.
A negative Joule-Thomson coefficient (\mu_{JT} < 0) indicates heating upon expansion.

Inversion Temperature

The temperature at which the Joule-Thomson coefficient changes sign is called the inversion temperature (T_i).
At the inversion temperature, \m$u_{JT} = 0$, meaning neither cooling nor heating occurs during the porous plug expansion.
For a Van der Waals gas, the inversion temperature depends on pressure and reaches a maximum inversion temperature T_{i,max} = \frac{2a}{bR}.

Behavior of Ideal vs. Real Gases

Ideal gases exhibit no intermolecular forces, resulting in a Joule-Thomson coefficient of zero (\m$u_{JT} = 0)$.
Real gases experience attractive forces at moderate temperatures and repulsive forces at high densities.
Hydrogen and helium show heating at ordinary room temperatures because their inversion temperatures are very low (around -80 °C for H_2 and -240 °C for He).

Derivations

Proof of Isenthalpic Nature of Porous Plug Experiment

Conditions: Steady flow, adiabatic enclosure, rigid non-conducting walls

Let 1 kg of gas pass from high pressure P_1, volume V_1 to low pressure P_2, volume V_2.
Work done on the gas by the high-pressure supply: $W_1 = P_1 V_1$.
Work done by the gas on the low-pressure region: $W_2 = P_2 V_2$.
Net work done by the gas on its surroundings: $W = P_2 V_2 - P_1 V_1$.
From the first law of thermodynamics for an adiabatic process ($Q = 0): \Delta U = -W \implies U_2 - U_1 = -(P_2 V_2 - P_1 V_1)$.
Rearranging yields U_1 + P_1 $V_1 = U_2 + P_2 V_2$, which means $H_1 = H_2$.

Result: $H_1 = H_2 (The porous plug expansion$ is an isenthalpic process)

Joule-Thomson Coefficient for a Van der Waals Gas

Conditions: Low pressures, Van der Waals gas

Write the Van der Waals equation: (P + \frac{a}{V^2})(V - b) = RT.
Expand and neglect the small term ab/V^2: PV + \frac{a}{V} - Pb = RT, or $V = \frac{RT}{P} + b - \frac{a}{RT}$.
Using the thermodynamic relation \mu_{JT} $C_p = T(\frac{\partial V}{\partial T})_P - V$.
Differentiate volume with respect to temperature at constant pressure: (\frac{\partial V}{\partial T})_$P = \frac{R}{P} + \frac{a}{RT^2}$.
Substitute into the coefficient equation: \mu_{JT} $C_p = T(\frac{R}{P} + \frac{a}{RT^2}) - (\frac{RT}{P} + b - \frac{a}{RT})$.
Simplify terms to obtain the final expression.

Result: \m$u_{JT} = \frac{1}{C_p}(\frac{2a}{RT} - b)$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Joule-Thomson Coefficient	$$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$$	T = Temperature, P = Pressure, H = Enthalpy K / Pa or K / atm	Isenthalpic process
First Law Flow Balance	$$U_1 + P_1V_1 = U_2 + P_2V_2$$	U = Internal energy, P = Pressure, V = Volume Joule (J)	Adiabatic steady flow
Enthalpy Conservation	$$H_1 = H_2$$	$H_1 = Initial enthalpy$, $H_2 = Final enthalpy$ Joule (J)	Porous plug expansion
Van der Waals Joule-Thomson Coefficient	$$\mu_{JT} = \frac{1}{C_p}\left(\frac{2a}{RT} - b\right)$$	$C_p = Molar heat capacity$, a, b = Van der Waals constants, R = Gas constant, T = Temperature K \cdot Pa^{-1}	Low pressure approximation
Maximum Inversion Temperature	$$T_{i,max} = \frac{2a}{bR}$$	a, b = Van der Waals constants, R = Universal gas constant Kelvin (K)	Zero pressure limit
Thermodynamic Identity for \mu_{JT}	$$\mu_{JT} = - \frac{1}{C_p} \left[ V - T \left(\frac{\partial V}{\partial T}\right)_P \right]$$	$C_p = Heat capacity at constant pressure$, V = Volume, T = Temperature K / Pa	General thermodynamic relation
Inversion Temperature Condition	$$T_i = \frac{2a}{R(b + \frac{V_{cor}}{...})}$$	$T_i = Inversion temperature$ Kelvin (K)	Derived from \m$u_{JT} = 0$
Ideal Gas Joule-Thomson Coefficient	$$\mu_{JT} = 0$$	\m$u_{JT} = Joule-Thomson coefficient$ Dimensionless / SI	Ideal gas law holds

Graphs and experiments

Joule-Thomson Isenthalpic Curves (T-P Diagram)

Draw: Curves of temperature versus pressure for a constant enthalpy. A parabola-like boundary curve enclosing the region where \mu_{JT} > 0.

Interpret: The peak of each isenthalpic curve represents the inversion temperature at that pressure. Inside the inversion curve, cooling occurs (dT/dP > 0); outside, heating occurs.

Exam use: Tested via multiple choice questions regarding the region of cooling and heating on T-P plots.

Porous Plug Apparatus Setup

Draw: A tube with an adiabatic jacket, a porous plug in the center, manometers on both sides, and thermometers placed upstream and downstream.

Interpret: Measures the precise temperature drop or rise \Delta T across the plug for a known pressure drop \Delta P under steady adiabatic conditions.

Exam use: Conceptual questions on experimental design, heat insulation, and steady-state maintenance.

Variation of \mu_{JT} with Temperature

Draw: A straight line or decreasing curve crossing the horizontal temperature axis at $T = T_i$.

Interpret: Shows clearly that \mu_{JT} is positive below T_i (cooling), zero at T_i, and negative above T_i (heating).

Exam use: Interpreting graphs to find the inversion temperature for specific gases like Hydrogen or Nitrogen.

Solved numericals

TGT core

Why do hydrogen and helium gas show a rise in temperature (heating) when passed through a porous plug at ordinary room temperature?

Recall that the Joule-Thomson coefficient is given by \m$u_{JT} = \frac{1}{C_p}(\frac{2a}{RT} - b)$.
For hydrogen and helium, the attractive intermolecular force parameter 'a' is extremely small, making the term \frac{2a}{RT} smaller than the co-volume 'b' at room temperature.
Thus, \mu_{JT} becomes negative.
A negative Joule-Thomson coefficient implies heating upon pressure drop.

Answer: Due to extremely low inversion temperatures, \mu_{JT} < 0 at room temperature, causing heating.

PGT extension

Calculate the maximum inversion temperature for a gas whose Van der Waals constants are $a = 0.364 N m^4 / mol^2$ and $b = 4.2 \times 10^{-5} m^3 / mol$. Take $R = 8.314 J / mol K$.

Identify the formula for maximum inversion temperature: T_{i,max} = \frac{2a}{bR}.
Substitute the given values: a = 0.364, $b = 4.2 \times 10^{-5}$, R = 8.314.
Calculate numerator: 2 \times 0.364 = 0.728.
Calculate denominator: (4.2 \times 10^{-5}) \times $8.314 = 3.49188 \times 10^{-4}$.
Perform division: T_{i,max} = \frac{0.728}{3.49188 \times 10^{-4}} \approx 2084.8 K.

Answer: 2084.8 K

PGT extension

Find the Joule-Thomson coefficient for an ideal gas using thermodynamic relations.

Use the equation of state for an ideal gas: PV = RT \implies V = \frac{RT}{P}.
Substitute V into the thermodynamic identity: \m$u_{JT} = - \frac{1}{C_p} [V - T(\frac{\partial V}{\partial T})_P]$.
Evaluate the partial derivative: (\frac{\partial V}{\partial T})_P = \frac{R}{P}.
Multiply by T: T(\frac{\partial V}{\partial T})_P = \frac{RT}{P} = V.
Substitute back: \m$u_{JT} = - \frac{1}{C_p} [V - V] = 0$.

Answer: \m$u_{JT} = 0$

Assumptions and limitations

The process is strictly adiabatic; any heat leakage through the walls introduces significant error in temperature measurement.
The flow must be in a steady state so that kinetic energy changes between the wide and narrow parts of the tube are negligible.
Van der Waals approximations hold only at low to moderate pressures.

Common misconceptions

Common error: The porous plug experiment is an adiabatic expansion where work is done against a vacuum.

Correct: It is a throttling process against a constant opposing pressure, maintaining an isenthalpic condition (H = constant).

Unlike free expansion (Joule expansion) where internal energy remains constant, throttling involves constant enthalpy.

Common error: All gases cool down when expanded through a porous plug.

Correct: Gases cool only when their temperature is below their inversion temperature.

Above the inversion temperature, real gases heat up because repulsive forces dominate.

Common error: An ideal gas experiences Joule-Thomson cooling.

Correct: Ideal gases show zero Joule-Thomson effect.

Absence of intermolecular forces means internal energy depends solely on temperature, and enthalpy depends only on temperature.

Quick revision

Porous plug experiment is an isenthalpic (H = constant) throttle process.
Joule-Thomson coefficient \m$u_{JT} = (\partial T / \partial P)_H determines cooling (\mu > 0) or heating (\mu < 0)$.
Inversion temperature is the temperature where \m$u_{JT} = 0$.
For ideal gases, \m$u_{JT} = 0 always$.
Maximum inversion temperature for a Van der Waals gas is given by T_{i,max} = 2a / (bR).

Must remember

The porous plug experiment is an isenthalpic process where $H_1 = H_2$.
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 $\mu_{JT} = 0$, meaning no cooling or heating occurs.
The maximum inversion temperature for a Van der Waals gas is given by $T_{i,max} = \frac{2a}{bR}$.
Hydrogen and helium show heating at ordinary room temperatures because their inversion temperatures are exceptionally low.

Glossary

Joule-Thomson Coefficient ($\mu_{JT}$)The rate of change of temperature with pressure at constant enthalpy, indicating whether a gas cools or heats upon expansion.

Inversion Temperature ($T_i$)The temperature at which the Joule-Thomson coefficient changes sign and $\mu_{JT} = 0$.

Isenthalpic ProcessA thermodynamic process in which the enthalpy of the system remains constant ($H_1 = H_2$).

Throttling ProcessA steady-flow adiabatic expansion through a porous plug or a restricted opening against a constant pressure.

Ideal Gas vs. Real Gas in Joule-Thomson Effect

Property	Ideal Gas	Real Gas
Joule-Thomson Coefficient ($\mu_{JT}$)	Zero ($\mu_{JT} = 0$)	Non-zero (depends on $T$ and $P$)
Intermolecular Forces	None	Present (attractive and repulsive)
Effect of Expansion	No temperature change	Cooling or heating depending on inversion temperature

Practice Quiz

Choose an option to reveal the explanation.

1. What thermodynamic property remains constant during the porous plug experiment?

Explanation: The porous plug experiment is a throttling process that is steady-flow and adiabatic, resulting in an isenthalpic process where initial enthalpy equals final enthalpy ($H_1 = H_2$).

2. How is the Joule-Thomson coefficient $\mu_{JT}$ mathematically defined?

Explanation: The Joule-Thomson coefficient is defined as the rate of change of temperature with pressure at constant enthalpy: $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$.

3. What does a positive Joule-Thomson coefficient ($\mu_{JT} > 0$) signify?

Explanation: A positive Joule-Thomson coefficient ($\mu_{JT} > 0$) implies that as pressure decreases ($\Delta P < 0$), temperature decreases ($\Delta T < 0$), resulting in cooling.

4. What is the Joule-Thomson coefficient for an ideal gas?

Explanation: Ideal gases have no intermolecular forces, meaning their internal energy and enthalpy depend only on temperature, giving $\mu_{JT} = 0$.

5. Which of the following gases shows heating at ordinary room temperatures during a porous plug expansion?

Explanation: Hydrogen and helium have very low inversion temperatures (around $-80^{\circ}\text{C}$ for $\text{H}_2$ and $-240^{\circ}\text{C}$ for $\text{He}$), so at ordinary room temperatures they are above their inversion temperatures and exhibit heating.

6. What is the formula for the maximum inversion temperature of a Van der Waals gas?

Explanation: For a Van der Waals gas, the maximum inversion temperature derived from the low pressure approximation and $\mu_{JT} = 0$ is $T_{i,max} = \frac{2a}{bR}$.

7. Which relation gives the Van der Waals Joule-Thomson coefficient?

Explanation: The Van der Waals Joule-Thomson coefficient at low pressures is given by $\mu_{JT} = \frac{1}{C_p}\left(\frac{2a}{RT} - b\right)$.

8. At the inversion temperature, what is the value of the Joule-Thomson coefficient?

Explanation: By definition, the inversion temperature $T_i$ is the temperature at which the Joule-Thomson coefficient changes sign, meaning $\mu_{JT} = 0$.

9. Why do real gases experience heating above their inversion temperature?

Explanation: Above the inversion temperature, the kinetic energy effects and repulsive molecular forces dominate, causing the gas to warm up upon expansion.

10. What is the general thermodynamic identity for $\mu_{JT}$?

Explanation: The general thermodynamic identity relating $\mu_{JT}$ to expansivity and heat capacity is $\mu_{JT} = - \frac{1}{C_p} \left[ V - T \left(\frac{\partial V}{\partial T}\right)_P \right]$.

11. What condition applies to the first law flow balance in a porous plug experiment?

Explanation: For steady-flow adiabatic throttling through a porous plug, $U_1 + P_1V_1 = U_2 + P_2V_2$, which means $H_1 = H_2$.

12. What is the approximate inversion temperature of hydrogen gas at ordinary conditions?

Explanation: Hydrogen shows heating at ordinary room temperatures because its inversion temperature is very low, around $-80^{\circ}\text{C}$.

13. What is the approximate inversion temperature of helium gas?

Explanation: Helium has an extremely low inversion temperature of around $-240^{\circ}\text{C}$, causing it to heat up upon expansion at room temperature.

14. Which units are appropriate for the Joule-Thomson coefficient?

Explanation: Since $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$, its units are temperature change per unit pressure change, i.e., $\text{K} / \text{Pa}$ or $\text{K} / \text{atm}$.

15. In the Joule-Thomson effect, what causes cooling in real gases below their inversion temperature?

Explanation: Below the inversion temperature, gas molecules expand against attractive intermolecular forces, doing internal work that decreases kinetic energy and results in cooling.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. For a real gas obeying the Van der Waals equation, the condition for inversion temperature where $\mu_{JT} = 0$ is given by:

Explanation: Setting the Vander Waals Joule-Thomson coefficient equation $\mu_{JT} = \frac{1}{C_p}\left(\frac{2a}{RT} - b\right) = 0$ yields $T_{i,max} = \frac{2a}{bR}$ at zero pressure limit.

2. Which of the following statements is correct regarding the porous plug experiment?

Explanation: The porous plug experiment forces a gas through a plug in an adiabatically walled tube against a constant low pressure, maintaining a steady-flow isenthalpic process.

3. Why does hydrogen gas show a rise in temperature when expanded through a porous plug at room temperature?

Explanation: Hydrogen has an inversion temperature of about $-80^{\circ}\text{C}$. At room temperature, it is above its inversion temperature, so $\mu_{JT} < 0$, leading to heating.

4. The Joule-Thomson coefficient $\mu_{JT}$ for an ideal gas is:

Explanation: Since ideal gases lack intermolecular forces, internal energy and enthalpy are functions of temperature alone, resulting in $\mu_{JT} = 0$.

5. In the throttling process, which of the following quantities remains unchanged?

Explanation: A throttling process is designed to be adiabatic and steady-flow, ensuring that the enthalpy before and after the valve/plug is identical ($H_1 = H_2$).

6. If $\mu_{JT} < 0$ for a gas during expansion, the temperature of the gas will:

Explanation: Since $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$, a negative coefficient means that as pressure decreases ($\Delta P < 0$), temperature increases ($\Delta T > 0$), causing heating.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the primary distinction between Joule expansion (free expansion) and Joule-Thomson expansion (throttling)?

Explanation: Joule expansion is free expansion into a vacuum where $U = \text{constant}$, whereas Joule-Thomson throttling is against a constant opposing pressure where $H = \text{constant}$.

2. Using the thermodynamic identity $\mu_{JT} = - \frac{1}{C_p} \left[ V - T \left(\frac{\partial V}{\partial T}\right)_P \right]$, what does the term $\left(\frac{\partial V}{\partial T}\right)_P$ represent?

Explanation: The partial derivative $\left(\frac{\partial V}{\partial T}\right)_P$ relates to thermal expansion and volume changes with temperature at constant pressure.

3. At what condition does a real gas transition from cooling to heating during a Joule-Thomson expansion?

Explanation: The inversion temperature is precisely the threshold where $\mu_{JT} = 0$, separating the cooling region (below $T_i$) from the heating region (above $T_i$).

4. What happens to the pressure and enthalpy of a gas when it passes through a porous plug?

Explanation: Gas is forced from a high-pressure region to a low-pressure region (pressure decreases), while the adiabatic steady flow ensures enthalpy remains constant.

5. Which gas parameter in the Van der Waals equation primarily accounts for molecular attractive forces in relation to the inversion temperature?

Explanation: The Van der Waals constant $a$ accounts for intermolecular attraction, which appears in the numerator of the maximum inversion temperature formula $T_{i,max} = \frac{2a}{bR}$.

6. Why does helium exhibit heating when expanded through a porous plug at room temperature?

Explanation: Helium's inversion temperature is approximately $-240^{\circ}\text{C}$. Room temperature is well above this value, placing helium in the heating regime.

7. In terms of work done, what occurs during steady-flow throttling across a porous plug?

Explanation: High-pressure gas does work on the plug/surroundings ($P_1V_1$), and work is done on the gas by the low-pressure side ($P_2V_2$), leading to the isenthalpic balance $U_1 + P_1V_1 = U_2 + P_2V_2$.

8. If a gas is at a temperature below its inversion temperature, what occurs during throttling?

Explanation: When $T < T_i$, the Joule-Thomson coefficient $\mu_{JT} > 0$, which causes the temperature to drop upon expansion (cooling).

9. What is the physical significance of $b$ in the Van der Waals expression for the Joule-Thomson coefficient $\mu_{JT} = \frac{1}{C_p}\left(\frac{2a}{RT} - b\right)$?

Explanation: The Van der Waals constant $b$ represents the co-volume or excluded volume per mole, accounting for the repulsive forces at high densities.

10. Can an ideal gas ever show a Joule-Thomson cooling effect?

Explanation: Because ideal gases possess zero intermolecular forces, their enthalpy depends solely on temperature, resulting in $\mu_{JT} = 0$ under all conditions.