Interactive physics simulator
Isothermal Process: PV = Constant
Explore thermodynamic transitions at a constant temperature. Observe how gas cylinders and syringes exchange heat with water baths to maintain equilibrium during expansion and compression.
Isothermal Process Lab
Set bath temperatures and slow plunger actions to study the relationship between pressure, volume, and heat transfer.
Live Telemetry
- Gas Temp (T)
- 300.0 K
- Pressure (P)
- 101.3 kPa
- Volume (V)
- 3.00 L
- Product (P·V)
- 303.9 N·m
- Work Done (W)
- 0.0 J
- Heat Flow (Q)
- 0.0 J
- Change (ΔU)
- 0.0 J
- Bath Temp
- 320.0 K
- Cylinder Press
- 101.3 kPa
- Volume (V)
- 40.0 mL
- Constant (PV)
- 4.05
- Work Done (W)
- 0.00 J
- Heat Input (Q)
- 0.00 J
- Jacket Temp
- 293.0 K
- Pressure (P)
- 100.0 kPa
- Volume (V)
- 3.00 L
- Work Input (W)
- -0.0 J
- Heat Removed (Q)
- -0.0 J
- Change (ΔU)
- 0.0 J
What is an Isothermal Process?
An isothermal process is a thermodynamic process in which the temperature of the system remains completely constant throughout the entire transition (T = constant). For an ideal gas, because the temperature does not change, the average kinetic energy of the gas molecules remains constant, meaning the total internal energy of the gas is unchanged:
Applying the First Law of Thermodynamics (ΔU = Q - W), this condition yields a direct equivalence between heat exchange and work:
During an isothermal expansion, as the gas does work on its surroundings, heat must flow into the system to prevent it from cooling. During an isothermal compression, as work is performed on the gas, heat must flow out of the system to prevent it from warming up.
Practical Conditions for T = Constant
Maintaining constant temperature during compression or expansion is challenging. To achieve it:
- Highly Conductive Walls: The gas container must have thin, highly conductive metal walls (such as copper) that facilitate quick heat transfer.
- Large Thermal Reservoir: The system must be submerged in a constant-temperature bath (like a water bath) that can absorb or release heat without altering its own temperature.
- Extremely Slow Process: The piston must move very slowly. This provides sufficient time for thermal energy to flow across the boundaries, maintaining temperature equilibrium at all times.
Boyle's Law & State Equations
For an ideal gas, pressure (P), volume (V), temperature (T), and amount (n) are related by:
Since the temperature T, moles n, and gas constant R are all constant, the right side is constant. Thus:
This is **Boyle's Law**. If the volume of the gas is doubled, its pressure is exactly halved, and vice-versa, satisfying:
Deriving Isothermal Work
Thermodynamic work is defined as W = ∫ P dV. For an isothermal ideal gas, we substitute P = (nRT) / (V):
Since n, R, and T are constant, they can be pulled out of the integral:
Evaluating the integral yields the natural logarithm:
Solved Examples
An ideal gas is held inside a cylinder at a constant temperature of 300 K. The gas expands isothermally from an initial volume of 2.0 Liters to a final volume of 6.0 Liters. If the gas contains 0.25 moles of molecules, calculate the mechanical work done by the gas during this expansion.
- Identify the given values: Temperature, T = 300 K. Initial volume, Vi = 2.0 L = 2.0 × 10-3 m3. Final volume, Vf = 6.0 L = 6.0 × 10-3 m3. Gas amount, n = 0.25 moles. Universal gas constant, R = 8.314 J/(mol·K).
- Recall the isothermal work formula: W = nRT ln(Vf / Vi).
- Calculate the volume ratio: Vf / Vi = 6.0 / 2.0 = 3.0.
- Substitute values into the equation: W = 0.25 · 8.314 · 300 · ln(3.0).
- Compute the initial product: 0.25 · 8.314 · 300 = 623.55 Joules.
- Calculate the natural log term: ln(3.0) ≈ 1.0986.
- Multiply the terms to find work done: W = 623.55 · 1.0986 ≈ 685 Joules.
- Interpret the result: The work is positive (+685 J), indicating that the gas does work on its surroundings by expanding.
Answer: W = +685 J
During an isothermal compression, 500 Joules of mechanical work is performed on 1.2 moles of an ideal gas at a constant temperature of 350 K. Calculate: (a) the change in internal energy (ΔU) of the gas, and (b) the net heat exchange (Q) with the constant-temperature water bath.
- Identify the constant temperature: T = 350 K.
- For an ideal gas, internal energy depends solely on temperature: U = U(T). Since the process is isothermal (ΔT = 0), the change in internal energy is exactly zero: ΔU = 0 Joules.
- Recall the First Law of Thermodynamics: ΔU = Q - W.
- Substitute ΔU = 0: 0 = Q - W, which implies Q = W.
- Identify the sign of work: Work is performed ON the gas (compression), meaning work done by the system is negative: W = -500 J.
- Determine heat exchange: Q = W = -500 Joules.
- Explain the sign: The negative sign for heat (-500 J) indicates that 500 Joules of heat energy flowed out of the gas and into the surrounding water bath to maintain the constant temperature of 350 K.
Answer: (a) ΔU = 0 J, (b) Q = -500 J (Heat released to bath)
A sample of gas undergoes an isothermal process where its volume is reduced to exactly one-half of its initial volume (Vf = 0.5 · Vi). If the initial pressure of the gas was 101.3 kPa, calculate the final pressure of the gas.
- Identify the process: Since temperature is constant, this is an isothermal process for an ideal gas.
- Recall Boyle's Law (isothermal equation of state): P1 · V1 = P2 · V2.
- Substitute the final volume expression: Pi · Vi = Pf · (0.5 · Vi).
- Cancel the volume term Vi from both sides: Pi = 0.5 · Pf.
- Solve for the final pressure: Pf = Pi / 0.5 = 2 · Pi.
- Perform the calculation: Pf = 2 · 101.3 kPa = 202.6 kPa.
- Conclude the result: Compressing the gas to half its volume at constant temperature exactly doubles its pressure.
Answer: Pf = 202.6 kPa
Common Mistakes
- Confusing Isothermal and Adiabatic: Thinking they are the same because they both represent gas expansions. Isothermal requires a thermal bath (Q ≠ 0, T = constant). Adiabatic requires thermal insulation (Q = 0, T changes).
- Assuming Delta U is non-zero: Calculating an internal energy change using constant volume or constant pressure formulas. For an ideal gas, because T is constant, ΔU is strictly zero.
- Log base error in work: Using common logarithm (log10) instead of natural logarithm (ln or loge) in calculations. Always use `ln`.
- Incorrect units for R: Using the gas constant R = 0.0821 L·atm/(mol·K) without converting pressure to atm and volume to Liters. Use R = 8.314 J/(mol·K) if pressure is in Pascals and volume is in cubic meters.
Microscopic Heat-Work Transmutation
At the molecular scale, adding heat to a gas increases the kinetic energy of its molecules, causing them to move faster.
However, in an isothermal expansion, as the piston moves outward, molecules collide with a moving boundary. Each collision with the retreating boundary causes a molecule to lose speed and bounce back slower, which cools the gas. The heat entering from the bath continuously replaces this lost molecular speed, keeping the average kinetic energy (and temperature) perfectly constant.
Practice Questions
1. What are the two essential physical conditions required to carry out a process isothermally in practice?
First, the container walls must have high thermal conductivity (such as thin copper rather than insulated plastic) to allow rapid heat exchange. Second, the expansion or compression must occur extremely slowly. This gives the gas enough time to exchange heat with the surroundings, maintaining thermal equilibrium.
2. For an ideal gas undergoing an isothermal expansion, explain where the energy comes from to perform expansion work, since its internal energy remains constant.
Since the process is isothermal, the internal energy of the ideal gas does not change (ΔU = 0). Under the First Law of Thermodynamics (ΔU = Q - W = 0), we find Q = W. This means that the mechanical work (W) done by the expanding gas to push the piston is completely powered by the thermal energy (Q) absorbed from the water bath.
3. Why is the P-V curve of an isothermal process (an isotherm) shallower than the curve of an adiabatic process starting from the same initial state?
In an isothermal expansion, heat enters the gas to maintain its temperature, keeping its pressure relatively high. In an adiabatic expansion, no heat enters, so temperature drops, causing pressure to drop much faster. Mathematically, the isothermal slope is dP/dV = -P/V, while the adiabatic slope is steeper: dP/dV = -γP/V (where γ ≈ 1.4).
4. Explain why a phase transition (like water boiling at 100°C) is considered an isothermal process, even though heat is added and volume changes.
During boiling, the temperature remains locked at 100°C because the added thermal energy (latent heat) is not used to speed up the molecules (which would raise temperature). Instead, it is used to break intermolecular hydrogen bonds, changing the molecular potential energy. Since temperature stays constant, it is an isothermal process.
FAQ
Frequently Asked Questions
What is an isothermal process?
An isothermal process is a thermodynamic process in which the temperature of the system remains completely constant throughout (T = constant). This is typically achieved by keeping the system in thermal contact with a large heat reservoir.
What is the formula for work done in an isothermal process?
For n moles of an ideal gas expanding or compressing isothermally from volume V_initial to V_final at absolute temperature T, the work done is given by W = nRT ln(Vf / Vi), where R is the universal gas constant.
What is the Ideal Gas Law relation in an isothermal process?
Since temperature is constant, the ideal gas equation PV = nRT simplifies to PV = constant (Boyle's Law). For any two states in the process, P1V1 = P2V2.
What is the change in internal energy during an isothermal process?
For an ideal gas, internal energy depends solely on temperature. Therefore, during an isothermal process, the change in internal energy is exactly zero (ΔU = 0).
Why does heat need to be added to a gas during isothermal expansion?
During expansion, the gas performs positive work on its surroundings, which removes energy from the gas molecules. To prevent the temperature from dropping, an equivalent amount of heat (Q = W) must flow from the surroundings into the gas.
Why must isothermal processes occur very slowly?
They must occur slowly to allow sufficient time for heat transfer across the system boundaries, ensuring the system remains in continuous thermal equilibrium with the heat bath.
What does the area under an isotherm on a P-V diagram represent?
The geometric area under the isotherm curve on a Pressure-Volume (P-V) diagram represents the total work done by the gas (or on the gas) during the process.
What is the difference between an isothermal process and an adiabatic process?
In an isothermal process, temperature remains constant and heat is freely exchanged with the surroundings (Q ≠ 0). In an adiabatic process, the system is thermally insulated so no heat is exchanged (Q = 0), and temperature changes during expansion or compression.
Can a process be isothermal if phase changes occur?
Yes. Phase changes like melting ice at 0°C or boiling water at 100°C are isothermal processes. The temperature remains constant while heat is added, and the energy goes entirely into breaking intermolecular bonds instead of changing temperature.
What is the sign of work and heat during isothermal compression?
During compression, the volume decreases (V_final < V_initial), so the work done by the gas is negative (W < 0) and heat flows out of the gas into the reservoir (Q < 0) to prevent the gas from warming up.