Interactive Lecture
Isothermal, Isobaric, Isochoric, and Adiabatic Processes
by Dr Arnaldo Delli Carri
In thermodynamics, a process is a transformation of a system from one equilibrium state to another. For an ideal gas, the relationship between pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the number of moles (\(n\)) is governed by the ideal gas law:
\[PV = nRT\]
Where:
How a gas behaves during a process determines how efficiently you can extract work from it or put work into it. Every engine, compressor, turbine, and refrigeration system is built around these four ideal processes:
An isothermal process occurs when the temperature remains constant throughout the transformation. This requires perfect thermal contact with a heat reservoir.
Key Relations:
\[T = \text{constant}\]
\[PV = \text{constant}\]
\[P_1V_1 = P_2V_2\]
Work Done: \[W = nRT\ln\left(\frac{V_2}{V_1}\right) = P_1V_1\ln\left(\frac{V_2}{V_1}\right)\]
Heat Transfer: \[Q = W\]
Internal Energy Change: \[\Delta U = 0\]
All the heat the gas absorbs during isothermal expansion leaves again as work. The particles keep the same average kinetic energy, so temperature is unchanged; the energy that pushes the piston is drawn straight from the reservoir.
An isobaric process occurs at constant pressure. This is common in systems with movable pistons exposed to atmospheric pressure.
Key Relations:
\[P = \text{constant}\]
\[\frac{V}{T} = \text{constant}\]
\[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]
Work Done: \[W = P(V_2 - V_1) = nR(T_2 - T_1)\]
Heat Transfer: \[Q = nC_P\Delta T\]
Internal Energy Change: \[\Delta U = nC_V\Delta T\]
Heat added at constant pressure splits two ways. The gas expands in proportion to temperature, so part of the heat raises internal energy and part does expansion work on the surroundings.
An isochoric process (also called isovolumetric) occurs when the volume remains constant. No boundary work is done in this process.
Key Relations:
\[V = \text{constant}\]
\[\frac{P}{T} = \text{constant}\]
\[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]
Work Done: \[W = 0\]
Heat Transfer: \[Q = nC_V\Delta T\]
Internal Energy Change: \[\Delta U = Q = nC_V\Delta T\]
With the volume fixed, every joule of heat becomes internal energy. The particles speed up and temperature rises, but with no piston movement the gas does no work.
An adiabatic process occurs with no heat exchange between the system and surroundings. This requires perfect thermal insulation or a very rapid process.
Key Relations:
\[Q = 0\]
\[PV^\gamma = \text{constant}\]
\[TV^{\gamma-1} = \text{constant}\]
\[P_1V_1^\gamma = P_2V_2^\gamma\]
Where \(\gamma = \frac{C_P}{C_V}\) is the heat capacity ratio
Work Done: \[W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}\]
Internal Energy Change: \[\Delta U = -W\]
With no heat allowed in or out, the gas pays for expansion work out of its own internal energy and cools. Compress it instead and the work done on the gas raises internal energy, so it heats up.
All these processes obey the First Law of Thermodynamics, which is the conservation of energy:
\[\Delta U = Q - W\]
Where:
The heat capacity of an ideal gas depends on whether heating occurs at constant volume or constant pressure:
Constant Volume: \[C_V = \left(\frac{\partial U}{\partial T}\right)_V\]
Constant Pressure: \[C_P = C_V + R\]
Heat Capacity Ratio: \[\gamma = \frac{C_P}{C_V}\]
For ideal gases:
Entropy \(S\) measures how widely a system's energy is spread among its accessible microstates. For a reversible change it is defined through \(dS = \delta Q_\text{rev}/T\), so its units are J/K. For an ideal gas the change between two states is:
\[\Delta S = nC_V\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}\]
On a temperature-entropy (T-S) diagram each process traces its own path: an isothermal is a horizontal line, an isochoric and an isobaric are logarithmic curves, and a reversible adiabatic is a vertical line because \(Q = 0\) forces \(\Delta S = 0\). The isochoric is the steeper of the two log curves: the slope is \(dT/dS = T/(nC)\), so the smaller heat capacity \(C_V\) gives the steeper rise. The simulation's T-S plot uses normalised entropy units for clarity.
Use the controls below to explore different thermodynamic processes. The left panel shows gas particles in a piston-cylinder, while the right panel displays the process on a thermodynamic diagram.
| Process | Constraint | Path Equation | Work Done | Heat Transfer | \(\Delta U\) |
|---|---|---|---|---|---|
| Isothermal | \(T = \text{const}\) | \(PV = \text{const}\) | \(nRT\ln\frac{V_2}{V_1}\) | \(W\) | \(0\) |
| Isobaric | \(P = \text{const}\) | \(\frac{V}{T} = \text{const}\) | \(P(V_2 - V_1)\) | \(nC_P\Delta T\) | \(nC_V\Delta T\) |
| Isochoric | \(V = \text{const}\) | \(\frac{P}{T} = \text{const}\) | \(0\) | \(nC_V\Delta T\) | \(Q\) |
| Adiabatic | \(Q = 0\) | \(PV^\gamma = \text{const}\) | \(\frac{P_1V_1 - P_2V_2}{\gamma - 1}\) | \(0\) | \(-W\) |
An ideal gas at initial pressure 200 kPa and volume 5 L undergoes an isothermal expansion to a final volume of 15 L at temperature 350 K. Calculate:
(a) The final pressure
(b) The work done by the gas
(c) The heat transferred
Show Solution
Given:
(a) Final Pressure:
For isothermal process: \(P_1V_1 = P_2V_2\)
\[P_2 = \frac{P_1V_1}{V_2} = \frac{200 \times 5}{15} = 66.7 \text{ kPa}\]
(b) Work Done:
First, find number of moles:
\[n = \frac{P_1V_1}{RT} = \frac{200{,}000 \times 0.005}{8.314 \times 350} = 0.344 \text{ mol}\]
Work done (using \(nRT = P_1V_1\) avoids the rounding in \(n\)):
\[W = nRT\ln\frac{V_2}{V_1} = P_1V_1\ln 3 = 1000 \times 1.0986 = 1{,}099 \text{ J}\]
(c) Heat Transfer:
For isothermal process, \(\Delta U = 0\), so:
\[Q = W = 1{,}099 \text{ J}\]
Answer: (a) \(P_2 = 66.7\,\text{kPa}\); (b) \(W = 1099\,\text{J}\); (c) \(Q = 1099\,\text{J}\).
Air (assume \(\gamma = 1.4\)) at 100 kPa and 300 K is compressed adiabatically to one-quarter of its initial volume. Calculate:
(a) The final pressure
(b) The final temperature
(c) The work done per mole
Show Solution
Given:
(a) Final Pressure:
For adiabatic process: \(P_1V_1^\gamma = P_2V_2^\gamma\)
\[P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = 100 \times 4^{1.4} = 100 \times 6.964 = 696 \text{ kPa}\]
(b) Final Temperature:
Using \(T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}\):
\[T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times 4^{0.4} = 300 \times 1.741 = 522.3 \text{ K}\]
(c) Work Done per Mole:
For adiabatic compression (work done ON the gas is negative), carrying the unrounded \(T_2\):
\[W = \frac{R(T_1 - T_2)}{\gamma - 1} = \frac{8.314(300 - 522.3)}{0.4} = -4{,}621 \text{ J/mol}\]
The negative sign indicates work is done on the gas.
Answer: (a) \(P_2 = 696\,\text{kPa}\); (b) \(T_2 = 522\,\text{K}\); (c) \(W = -4621\,\text{J/mol}\).
A piston-cylinder device contains 2 mol of helium gas (monatomic, \(C_V = 12.5\) J/(mol·K)) at 150 kPa and 300 K. The gas is heated at constant pressure until its volume doubles. Calculate:
(a) The final temperature
(b) The work done
(c) The heat transferred
(d) The change in internal energy
Show Solution
Given:
(a) Final Temperature:
For isobaric process: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
\[T_2 = T_1\frac{V_2}{V_1} = 300 \times 2 = 600 \text{ K}\]
(b) Work Done:
\[W = P(V_2 - V_1) = nR(T_2 - T_1) = 2 \times 8.314 \times (600 - 300) = 4{,}988 \text{ J}\]
(c) Heat Transfer:
\(C_P = C_V + R = 12.5 + 8.314 = 20.814\) J/(mol·K)
\[Q = nC_P\Delta T = 2 \times 20.814 \times 300 = 12{,}488 \text{ J}\]
(d) Change in Internal Energy:
\[\Delta U = nC_V\Delta T = 2 \times 12.5 \times 300 = 7{,}500 \text{ J}\]
Verify: \(\Delta U = Q - W = 12{,}488 - 4{,}988 = 7{,}500\) J ✓
Answer: (a) \(T_2 = 600\,\text{K}\); (b) \(W = 4988\,\text{J}\); (c) \(Q = 12{,}488\,\text{J}\); (d) \(\Delta U = 7500\,\text{J}\).
A rigid container holds 0.5 mol of nitrogen gas (diatomic) at 250 K and 80 kPa. Heat is added until the pressure reaches 200 kPa. For nitrogen, \(C_V = 20.8\) J/(mol·K). Calculate:
(a) The final temperature
(b) The heat transferred
(c) The change in internal energy
Show Solution
Given:
(a) Final Temperature:
For isochoric process: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
\[T_2 = T_1\frac{P_2}{P_1} = 250 \times \frac{200}{80} = 625 \text{ K}\]
(b) Heat Transfer:
For constant volume process:
\[Q = nC_V\Delta T = 0.5 \times 20.8 \times (625 - 250) = 3{,}900 \text{ J}\]
(c) Change in Internal Energy:
Since \(W = 0\) for isochoric process:
\[\Delta U = Q - W = 3{,}900 - 0 = 3{,}900 \text{ J}\]
Answer: (a) \(T_2 = 625\,\text{K}\); (b) \(Q = 3900\,\text{J}\); (c) \(\Delta U = 3900\,\text{J}\).