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Isobaric Process: P = Constant

Explore thermodynamic cycles at constant pressure. Discover how adding or removing heat directly translates to expansion work (W = PΔV) and temperature shifts in physical setups.

Isobaric Process Lab ⓘ

Alter heat input rates, gas amounts, and initial temperatures to study Charles's Law (V/T = Constant).

Simulating...

Live Telemetry

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

What is an Isobaric Process?

An isobaric process is a thermodynamic process in which the pressure of the system remains completely constant throughout the transition (ΔP = 0). The term comes from the Greek words isos (equal) and baros (weight/pressure).

For an ideal gas, because the boundaries are free to expand or contract (like a weighted piston), the volume expands as temperature rises, following Charles's Law:

V / T = constant

According to the First Law of Thermodynamics (ΔU = Q - W), the heat added (Q) is spent on both increasing the internal energy (ΔU) and performing mechanical expansion work (W = PΔV) against the environment:

Q = ΔU + P·ΔV

Governing Equations for Isobaric Processes

For n moles of an ideal gas undergoing a reversible constant-pressure process:

Charles's Law

V1 / T1 = V2 / T2

Volume increases linearly with absolute temperature. Plotted as a diagonal straight line on a V-T diagram.

Mechanical Work Done

W = P · ΔV = nR · ΔT

Work done equals pressure times change in volume. Represented as a flat rectangle under the isobaric path on a P-V graph.

Heat Capacity (Q)

Q = n · Cp · ΔT

Cp is the molar heat capacity at constant pressure. Cp = Cv + R, because energy is also consumed doing work.

Solved Examples

Example 1

A gas undergoes an isobaric expansion at a constant pressure of 150 kPa. During this process, the volume of the gas increases from 2.0 Liters to 5.0 Liters. If 750 Joules of heat energy are added to the gas during the expansion, 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: Constant pressure, P = 150 kPa = 150 × 103 Pa. Initial volume, Vi = 2.0 L = 2.0 × 10-3 m3. Final volume, Vf = 5.0 L = 5.0 × 10-3 m3. Heat added, Q = +750 Joules.
  2. Calculate the change in volume: ΔV = Vf - Vi = 5.0 L - 2.0 L = 3.0 L = 3.0 × 10-3 m3.
  3. Use the isobaric work formula: W = P · ΔV.
  4. Substitute values into the work formula: W = (150 × 103 Pa) · (3.0 × 10-3 m3) = 450 Joules. (The gas does positive mechanical work on the surroundings.)
  5. Apply the First Law of Thermodynamics: ΔU = Q - W.
  6. Substitute the heat and work values to find internal energy change: ΔU = 750 J - 450 J = 300 Joules. (The internal energy increases by 300 J, causing the gas temperature to rise.)
Final Answer: Work Done, W = 450 J; Internal Energy Change, ΔU = +300 J
Example 2

A sample of 0.20 moles of dry air (a diatomic gas with C<sub>p</sub> = 29.1 J/(mol&middot;K) and &gamma; = 1.40) is heated at a constant pressure of 101.3 kPa from an initial temperature of 300 K to a final temperature of 450 K. Calculate: (a) the heat added to the air, (b) the work done by the air, and (c) the volume change. (Gas constant R = 8.314 J/(mol&middot;K)).

Step-by-Step Solution:

  1. Identify given values: moles, n = 0.20. Constant pressure, P = 101.3 kPa. Initial temp, Ti = 300 K. Final temp, Tf = 450 K. Specific heat, Cp = 29.1 J/(mol·K). Temperature change, ΔT = 450 K - 300 K = 150 K.
  2. Calculate heat added using specific heat at constant pressure: Q = n · Cp · ΔT.
  3. Substitute values: Q = 0.20 · 29.1 · 150 = 873 Joules.
  4. Recall specific heat relation Cv = Cp - R = 29.1 - 8.314 = 20.786 J/(mol·K). Calculate change in internal energy: ΔU = n · Cv · ΔT = 0.20 · 20.786 · 150 = 623.6 Joules.
  5. Use the First Law to solve for work: W = Q - ΔU = 873 J - 623.6 J = 249.4 Joules. Alternatively, use W = n · R · ΔT = 0.20 · 8.314 · 150 = 249.4 Joules. (Both methods yield identical results.)
  6. Solve for the volume change using ΔV = W / P: ΔV = 249.4 J / (101.3 × 103 Pa) = 2.46 × 10-3 m3 = 2.46 Liters.
Final Answer: (a) Q = 873 J, (b) W = 249.4 J, (c) ΔV = 2.46 L
Example 3

An open hot-air balloon containing 800 m<sup>3</sup> of air is heated at sea level (atmospheric pressure 101.3 kPa) so that its temperature rises from 290 K to 360 K. Because the envelope is open at the bottom, air expands and escapes into the atmosphere to maintain constant pressure. Calculate: (a) the final volume that the heated air would occupy if it were fully retained inside the expanding balloon, and (b) the expansion work done by the air parcel against the surrounding atmosphere.

Step-by-Step Solution:

  1. Identify the given values: Initial volume, Vi = 800 m3. Constant pressure, P = 101.3 kPa = 101.3 × 103 Pa. Initial temperature, Ti = 290 K. Final temperature, Tf = 360 K.
  2. Apply Charles's Law for isobaric expansion: Vi / Ti = Vf / Tf.
  3. Solve for final volume Vf: Vf = Vi · (Tf / Ti).
  4. Substitute values: Vf = 800 · (360 / 290) ≈ 800 · 1.2414 = 993.1 m3.
  5. Calculate the change in volume: ΔV = Vf - Vi = 993.1 m3 - 800 m3 = 193.1 m3.
  6. Calculate the expansion work done: W = P · ΔV = 101.3 × 103 Pa · 193.1 m3 = 19.56 × 106 Joules = 19.56 MJ. (This work represents the energy spent displacing atmospheric air, enabling buoyancy.)
Final Answer: (a) Vf ≈ 993.1 m3, (b) W = 19.56 MJ

Self-Check Practice Questions

Q1. Why is the specific heat capacity of a gas at constant pressure (Cp) always greater than its specific heat capacity at constant volume (Cv)?

Show Explanation
When a gas is heated at constant volume (Cv), no work is done (W = 0), so all added heat energy goes entirely into raising internal energy and temperature. When heated at constant pressure (Cp), the gas expands and does work on its surroundings (W = PΔV). Therefore, extra heat energy must be supplied to do this work in addition to raising the temperature, making Cp > Cv. The difference is Cp - Cv = R.

Q2. A balloon is placed inside a refrigerator. Explain what happens to its volume from a molecular perspective, assuming constant pressure.

Show Explanation
Inside the refrigerator, heat flows out of the balloon, lowering the temperature of the gas molecules. The average kinetic energy and velocity of the molecules decrease, reducing the force of their impacts on the balloon walls. To maintain constant pressure (balancing the atmospheric pressure outside), the balloon envelope collapses, reducing the volume (V ∝ T). The smaller volume keeps the frequency of wall collisions high enough to maintain constant pressure despite lower speeds.

Q3. Explain why the work done in an isobaric process is easy to compute on a P-V diagram compared to an isothermal or adiabatic process.

Show Explanation
On a Pressure-Volume (P-V) diagram, pressure is plotted on the Y-axis and volume on the X-axis. An isobaric process occurs at constant pressure, which draws a flat horizontal line. The area under a horizontal line from V_initial to V_final forms a perfect rectangle. The area of a rectangle is simply width times height, which corresponds directly to W = P · ΔV. Isothermal and adiabatic processes draw curved lines, requiring calculus integration.

Q4. Describe the role of the burner in a hot-air balloon in maintaining an isobaric process.

Show Explanation
The bottom of a hot-air balloon envelope is open to the atmosphere, meaning the pressure inside is always equal to the surrounding atmospheric pressure (constant pressure). When the burner fires, it heats the air inside. Since the pressure is fixed, the air expands adiabatically/isobarically. As it expands, hot air escapes through the open bottom. Since mass decreases while volume is constant, density decreases, providing buoyancy to lift the balloon.

Frequently Asked Questions

1. What is an isobaric process?

An isobaric process is a thermodynamic process in which the pressure of the system remains completely constant throughout the transition (ΔP = 0). This is typically achieved by allowing the boundaries of the system (like a piston) to expand or contract freely under a constant external force or weight.

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

For a constant pressure P undergoing a volume change from V_initial to V_final, the work done by the gas is given by W = P · ΔV = P · (Vfinal - Vinitial) = nR · (Tfinal - Tinitial).

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

Under the First Law (ΔU = Q - W), since both heat transfer and work done are non-zero, the heat added is split between changing the internal energy of the gas and doing mechanical work: Q = ΔU + W = nCp ΔT.

4. What is Charles's Law and how does it relate to isobaric processes?

Charles's Law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its absolute temperature: V / T = constant, or V1 / T1 = V2 / T2. This is the governing equation of state for an isobaric process.

5. What is the difference between Cp and Cv?

C_p is always larger than C_v because in a constant pressure process, some of the added heat must be spent on expansion work (W = PΔV) to keep pressure constant, whereas in a constant volume process, all heat goes entirely into raising temperature (ΔU). The relation is Cp - Cv = R.

6. Why does a gas expand when heated at constant pressure?

When a gas is heated, its molecules gain kinetic energy and collision speeds increase. To prevent the pressure (collision force per unit area) from rising, the gas must expand, increasing the volume and reducing the collision frequency to offset the higher velocities.

7. What does the area under an isobaric curve on a P-V diagram represent?

On a Pressure-Volume (P-V) diagram, an isobaric process is represented by a flat, horizontal line. The area under this horizontal line is a simple rectangle of height P and width ΔV, representing the work done W = P · ΔV.

8. What are some real-world examples of isobaric processes?

Examples include water boiling in an open container (where atmospheric pressure is constant), heating air inside a hot-air balloon causing it to expand and lift, and a piston-cylinder carrying a fixed mass that rises or falls as the gas is heated or cooled.

9. Is heat added during isobaric expansion positive or negative?

During isobaric expansion, temperature rises, meaning both internal energy increases (ΔU > 0) and expansion work is done by the gas (W > 0). Therefore, the heat added to the system is positive (Q > 0).

10. How is enthalpy related to an isobaric process?

In an isobaric process, the heat transferred into or out of the system is exactly equal to the change in the system's enthalpy (Q = ΔH), where enthalpy H is defined as H = U + PV.

First Law of ThermodynamicsWork Done by GasIsothermal ProcessAdiabatic ProcessCharles's Law