Use this simulation to identify isothermal, reversible-adiabatic and irreversible-adiabatic processes of an ideal gas in a step-by-step procedure. After starting and with “new problem” selected, the simulation shows either an expansion or a compression process, and either a pressure-temperature or pressure-volume diagram. Select your answer from the possible options (a, b, c, d, e) and then select “show solution” to see the correct answer. The “hint” button provides a hint for each step, and once “show solution” is selected, you cannot go back.
The first law of thermodynamics, representing the conservation of energy, is:
$\Delta U = Q - W$,
where $\Delta U$ is the change to internal energy of the system, $Q$ is heat added to the system and $W$ is the work done by the system. In adiabatic processes, $Q = 0$, while $Q \neq 0$ in isothermal processes with external pressure $P_{ext} \neq 0$. Expansion-compression work $W$ for all four processes is calculated from:
$W = -\int P_{ext} \, dV$,
where $P_{ext}$ is the external pressure and $W$ is in units of kJ/mol. The external pressure and the gas pressure are equal for a reversible process, whereas for an irreversible process the external pressure is equal to the final pressure. Change in internal energy is calculated from:
$\Delta U = C_V \, (T_2 - T_1)$.
Initial state:
$V_1 = \dfrac{R \, T_1}{P_1}$,
where the subscript $1$ refers to the initial state, $R$ is the ideal gas constant (kJ/mol K), $V$ is volume ($\mathrm{m}^3$/mol), $T$ is temperature (K) and $P$ is pressure (Pa).
For an isothermal process:
$V_2 = \dfrac{R \, T_1}{P_2}$,
where the subscript $2$ refers to the final condition.
Reversible work:
$W = -R \, T_1 \ln \left( \dfrac{V_2}{V_1} \right)$.
Irreversible work:
$W = -P_2 \, (V_2 - V_1)$.
For an adiabatic process on an ideal diatomic gas:
$\gamma = \dfrac{7}{5}$,
$C_V = \dfrac{5 R}{2}$,
$W = C_V \, (T_2 - T_1)$,
where $\gamma = C_p / C_V$, $C_V$ is the constant volume heat capacity and $C_p$ is the constant pressure heat capacity (kJ/(mol K)).
Reversible process:
$V_2 = V_1 \left( \dfrac{P_1}{P_2} \right)^{1/\gamma}$,
$T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$.
Irreversible process:
$T_2 = T_1 - \dfrac{P_2 \, (V_2 - V_1)}{C_V}$,
$V_2 = \dfrac{R \, (C_V T_1 + P_2 V_1)}{P_2 \, (C_V + R)}$.
Reference
[1] J. R. Elliott and C. T. Lira, Introductory Chemical Engineering Thermodynamics, 2nd ed., Upper Saddle River, NJ: Prentice Hall, 2012.
This simulation was created in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It is a JavaScript/HTML5 implementation of a Mathematica simulation by Neil Hendren. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988). Address any questions or comments to LearnChemE@gmail.com.