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Isochoric Process: V = Constant

Investigate thermodynamic processes at constant volume. Observe how heating a rigid container increases internal energy (Q = ΔU) and pressure without performing any mechanical work (W = 0).

Isochoric Process Lab ⓘ

Alter heat input rates, gas amounts, and initial temperatures to study Gay-Lussac's Law (P/T = Constant).

Simulating...

Live Telemetry

Gas Temp (T)
300.0 K
Pressure (P)
100.0 kPa
Volume (V)
5.00 L
Heat Input (Q)
0.0 J
Work Done (W)
0.0 J
Change (ΔU)
0.0 J

What is an Isochoric Process?

An isochoric process (also known as an isometric process or constant-volume process) is a thermodynamic process in which the volume of the closed system remains completely constant (ΔV = 0). The term is derived from the Greek words isos (equal) and chora (space/volume).

Because the volume is constant, the boundary of the system cannot expand or contract, which means the gas performs no boundary work on its surroundings:

W = P · ΔV = 0

According to the First Law of Thermodynamics (ΔU = Q - W), since work done is zero, the equation reduces to:

Q = ΔU

This signifies that all heat energy transferred into or out of the system is spent entirely on changing the internal thermal energy of the gas, leading directly to a change in temperature and pressure.

Governing Equations for Isochoric Processes

For n moles of an ideal gas undergoing a constant-volume transition:

Gay-Lussac's Law

P1 / T1 = P2 / T2

Pressure is directly proportional to absolute temperature. Plotted as a diagonal straight line passing through absolute zero on a P-T diagram.

Zero Mechanical Work

W = 0

No expansion or compression work occurs because the displacement of container boundaries is zero. Plotted as a vertical path on a P-V graph.

Molar Heat Capacity (Q)

Q = ΔU = n · Cv · ΔT

Cv is the molar heat capacity at constant volume. For monatomic gases, Cv = 1.5 R. For diatomic gases, Cv = 2.5 R.

Solved Examples

Example 1

A rigid glass flask containing a gas is heated. The pressure of the gas increases from 120 kPa to 240 kPa. If 800 Joules of heat energy are added to the gas during this heating process, calculate: (a) the work performed by the gas, and (b) the change in its internal energy.

Step-by-Step Solution:

  1. Identify the given values: Initial pressure, Pi = 120 kPa. Final pressure, Pf = 240 kPa. Heat added, Q = +800 Joules.
  2. Identify the process type: Since the flask is rigid, the volume remains completely constant throughout the process (ΔV = 0).
  3. Calculate the work done: In an isochoric process, since the boundary does not move, the work done by the gas is W = P · ΔV = P · 0 = 0 Joules.
  4. Apply the First Law of Thermodynamics: ΔU = Q - W.
  5. Substitute values to find internal energy change: ΔU = 800 J - 0 J = 800 Joules. (All added heat energy is converted entirely into internal thermal energy, raising the gas temperature.)
Final Answer: Work Done, W = 0 J; Internal Energy Change, ΔU = +800 J
Example 2

A sample of 0.15 moles of helium gas (a monatomic gas with C<sub>v</sub> = 12.47 J/(mol&middot;K)) is sealed inside a rigid metal cylinder. The gas is cooled from an initial temperature of 380 K to a final temperature of 280 K. Calculate: (a) the change in internal energy, (b) the work done by the gas, and (c) the heat energy transferred from the gas.

Step-by-Step Solution:

  1. Identify given values: moles, n = 0.15. Initial temp, Ti = 380 K. Final temp, Tf = 280 K. Specific heat, Cv = 12.47 J/(mol·K). Temperature change, ΔT = 280 K - 380 K = -100 K.
  2. Calculate the change in internal energy: ΔU = n · Cv · ΔT.
  3. Substitute values: ΔU = 0.15 · 12.47 · (-100) = -187.05 Joules. (The negative sign indicates the gas lost internal energy.)
  4. Identify work done: Since the cylinder is rigid, the volume change is zero (ΔV = 0), so work done W = 0 J.
  5. Apply the First Law of Thermodynamics to find heat transfer Q: Q = ΔU + W = -187.05 J + 0 J = -187.05 Joules. (The negative sign indicates that 187.05 J of heat was rejected to the surroundings.)
Final Answer: (a) ΔU = -187.05 J, (b) W = 0 J, (c) Q = -187.05 J (heat released)
Example 3

A compressed air tank with a fixed volume of 0.05 m³ is left in direct sunlight. During the day, the air temperature rises from 290 K to 348 K. If the initial pressure in the tank was 2.0 &times; 10⁵ Pa (200 kPa), calculate: (a) the final pressure inside the tank, and (b) the expansion work done by the gas.

Step-by-Step Solution:

  1. Identify given values: Fixed volume, V = 0.05 m³. Initial temp, Ti = 290 K. Final temp, Tf = 348 K. Initial pressure, Pi = 2.0 × 105 Pa.
  2. Apply Gay-Lussac's Law for constant volume: Pi / Ti = Pf / Tf.
  3. Solve for final pressure Pf: Pf = Pi · (Tf / Ti).
  4. Substitute values: Pf = (2.0 × 105 Pa) · (348 / 290) = 2.4 × 105 Pa = 240 kPa.
  5. Determine work done: Because the tank walls are rigid, the volume change is zero (ΔV = 0). Thus, the expansion work done is W = P · ΔV = 0 J.
Final Answer: (a) Final Pressure, Pf = 240 kPa (2.4 × 10⁵ Pa), (b) Work Done, W = 0 J

Self-Check Practice Questions

Q1. Why is no mechanical work performed during an isochoric process, even if the pressure increases significantly?

Show Explanation
Mechanical work done by a gas requires a change in volume ($W = \int P \, dV$). In an isochoric process, the boundaries of the container are completely rigid and fixed, meaning the change in volume is exactly zero ($\Delta V = 0$). Since there is no displacement against external pressure, no mechanical work is performed on the surroundings ($W = 0$).

Q2. Why does the temperature of a gas rise faster when heated at constant volume compared to constant pressure?

Show Explanation
When heated at constant volume ($Q = \Delta U$), all added heat energy goes entirely into increasing internal energy and temperature because no energy is spent doing work. When heated at constant pressure ($Q = \Delta U + W$), some of the heat must be used to perform expansion work ($P\Delta V$) against the atmosphere. Thus, more heat is required to produce the same temperature increase, making constant-volume heating faster.

Q3. Explain the safety hazard of exposing sealed aerosol spray cans or gas cylinders to open flames or direct sunlight.

Show Explanation
Aerosol cans and gas tanks are rigid containers with constant volume (isochoric setup). According to Gay-Lussac's Law ($P \propto T$), as the temperature of the container rises due to sunlight or fire, the internal pressure increases proportionally. Since the container cannot expand to relieve pressure, the pressure can exceed the structural strength of the metal walls, leading to a violent rupture or explosion.

Q4. What is the physical meaning of a vertical line on a Pressure-Volume (P-V) diagram?

Show Explanation
A vertical line on a P-V diagram indicates that the volume remains completely constant while the pressure changes. The area under a vertical line (the region between the path and the X-axis) is exactly zero because the width of the line is zero. This visually represents that the work done during the process is zero ($W = 0$).

Frequently Asked Questions

1. What is an isochoric process?

An isochoric process (also called a constant-volume process) is a thermodynamic process in which the volume of the system remains completely constant throughout the transition (ΔV = 0). This is achieved by enclosing the gas in a completely rigid container.

2. What is the formula for work done in an isochoric process?

Since volume is constant, the work done by the gas is exactly zero: W = P · ΔV = 0.

3. How does the First Law of Thermodynamics apply to an isochoric process?

Under the First Law (ΔU = Q - W), since W = 0, the equation simplifies to Q = ΔU. This means all heat energy added goes entirely into raising internal energy and temperature.

4. What is Gay-Lussac's Law?

Gay-Lussac's Law states that at constant volume, the pressure of a gas is directly proportional to its absolute temperature: P / T = constant, or P1 / T1 = P2 / T2.

5. Why does pressure rise when gas is heated at constant volume?

Heating increases the kinetic energy and velocity of gas molecules. Because the container volume is locked, the faster molecules collide with the container walls more frequently and with greater force, increasing pressure.

6. What does an isochoric process look like on a P-V diagram?

On a Pressure-Volume (P-V) diagram, an isochoric process is represented by a vertical straight line, indicating constant volume with changing pressure.

7. What is the difference between Cp and Cv?

C_p (heat capacity at constant pressure) is larger than C_v (heat capacity at constant volume) because at constant pressure, a gas must expand and do work, consuming additional heat energy, whereas at constant volume, no work is done.

8. What are some examples of isochoric processes?

Examples include heating a gas inside a sealed rigid steel cylinder, cooking inside a sealed pressure cooker, and autoclave sterilization where volume cannot change.

9. Is heat added during an isochoric process positive or negative?

If the temperature and internal energy of the gas increase, the heat added is positive (Q > 0). If the gas is cooled and temperature drops, heat is removed, making it negative (Q < 0).

10. How is enthalpy change related to an isochoric process?

Enthalpy is H = U + PV. For an isochoric process, volume is constant, so the enthalpy change is ΔH = ΔU + V·ΔP = Q + V·ΔP, meaning enthalpy change does not equal heat added due to the pressure rise.

First Law of ThermodynamicsWork Done by GasIsobaric ProcessIsothermal ProcessGay-Lussac's Law