Thermodynamics • Topic 2 of 3

Thermodynamic Processes

A thermodynamic process is the way a system passes from one state to another. By holding one quantity fixed we obtain four important reversible processes, each with its own $P$–$V$ signature and its own form of the work done $W=\int P\,dV$.

1. Isothermal process (constant temperature). The gas is kept at a fixed temperature, usually by surrounding it with a heat reservoir and changing the volume slowly. Since $T$ is constant, for an ideal gas $\Delta U=0$, and the first law gives $\Delta Q=\Delta W$ — all the heat goes into work (or all the work removes heat).

  • It obeys Boyle's law $PV=\text{constant}$, so the $P$–$V$ curve is a hyperbola (isotherm).
  • Work done: $W=nRT\ln\!\left(\frac{V_2}{V_1}\right)=2.303\,nRT\log_{10}\!\left(\frac{V_2}{V_1}\right)$.

2. Adiabatic process (no heat exchange). The system is thermally insulated, so $\Delta Q=0$. The first law then gives $\Delta W=-\Delta U$ — the gas does work entirely at the cost of its internal energy, so an adiabatic expansion cools the gas and a compression heats it.

  • It obeys $PV^\gamma=\text{const}$ and $TV^{\gamma-1}=\text{const}$, where $\gamma=\frac{C_p}{C_v}$.
  • The adiabatic curve is steeper than the isotherm on a $P$–$V$ diagram.
  • Work done: $W=\frac{P_1V_1-P_2V_2}{\gamma-1}=\frac{nR(T_1-T_2)}{\gamma-1}$.

3. Isobaric process (constant pressure). The pressure is held fixed while the volume changes (e.g. a gas heated under a free piston).

  • Work done: $W=P(V_2-V_1)=nR(T_2-T_1)$.
  • Heat supplied: $\Delta Q=nC_p\,\Delta T$, using the molar specific heat at constant pressure.

4. Isochoric process (constant volume). The volume is fixed, so $dV=0$ and the gas does no work: $W=0$. All the heat raises the internal energy: $\Delta Q=\Delta U=nC_v\,\Delta T$.

Specific heats and Mayer's relation. A gas has two principal molar specific heats: $C_v$ (at constant volume) and $C_p$ (at constant pressure). Heating at constant pressure needs extra heat to do expansion work, so $C_p>C_v$. The first law applied to one mole gives Mayer's relation:

  • $C_p-C_v=R$ (with $R=8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$).
  • Their ratio is the adiabatic exponent $\gamma=\frac{C_p}{C_v}$: about $1.67$ for monatomic and $1.40$ for diatomic gases.
P–V diagram: isothermal (gentle) versus adiabatic (steeper) expansionPV(P1,V1)Isothermal PV=constAdiabatic PV^g=constV1
Solved Examples
1
Worked Example
Two moles of an ideal gas expand isothermally at 300 K from a volume $V$ to $2V$. Find the work done. (Take $R=8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$.)
Solution
  1. Step 1: Isothermal work $W=nRT\ln\!\left(\frac{V_2}{V_1}\right)$.
  2. Step 2: Substitute $n=2$, $T=300$, $\frac{V_2}{V_1}=2$: $W=2\times8.314\times300\times\ln 2$.
  3. Step 3: With $\ln 2=0.693$: $W=4988.4\times0.693\approx3457\ \text{J}$.

Answer: $W\approx3.46\times10^{3}\ \text{J}$.

2
Worked Example
In an isothermal expansion a gas does 500 J of work. How much heat is absorbed by the gas?
Solution
  1. Step 1: For an isothermal process $T$ is constant, so $\Delta U=0$.
  2. Step 2: First law: $\Delta Q=\Delta U+\Delta W=0+\Delta W$.
  3. Step 3: Therefore $\Delta Q=\Delta W=500\ \text{J}$.

Answer: $\Delta Q=500\ \text{J}$ (all the absorbed heat becomes work).

3
Worked Example
The specific heat of a gas at constant volume is $C_v=20.8\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$. Find $C_p$ and $\gamma$. (Take $R=8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$.)
Solution
  1. Step 1: By Mayer's relation $C_p=C_v+R$.
  2. Step 2: Substitute: $C_p=20.8+8.314=29.1\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$.
  3. Step 3: $\gamma=\frac{C_p}{C_v}=\frac{29.1}{20.8}\approx1.40$ (a diatomic gas).

Answer: $C_p\approx29.1\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$, $\gamma\approx1.40$.

4
Worked Example
A gas at $P_1=8\times10^{5}\ \text{Pa}$, $V_1=2\times10^{-3}\ \text{m}^3$ expands adiabatically to $P_2=1\times10^{5}\ \text{Pa}$, $V_2=8\times10^{-3}\ \text{m}^3$. Take $\gamma=1.4$. Find the work done by the gas.
Solution
  1. Step 1: Adiabatic work $W=\frac{P_1V_1-P_2V_2}{\gamma-1}$.
  2. Step 2: $P_1V_1=8\times10^{5}\times2\times10^{-3}=1600\ \text{J}$; $P_2V_2=1\times10^{5}\times8\times10^{-3}=800\ \text{J}$.
  3. Step 3: $W=\frac{1600-800}{1.4-1}=\frac{800}{0.4}=2000\ \text{J}$.

Answer: $W=2000\ \text{J}$.

5
Worked Example
A gas expands at a constant pressure of $1.5\times10^{5}\ \text{Pa}$ and absorbs 1200 J of heat. Its volume increases by $4\times10^{-3}\ \text{m}^3$. Find the change in internal energy.
Solution
  1. Step 1: Isobaric work $W=P\,\Delta V=1.5\times10^{5}\times4\times10^{-3}=600\ \text{J}$.
  2. Step 2: First law: $\Delta U=\Delta Q-\Delta W=1200-600$.
  3. Step 3: Compute: $\Delta U=600\ \text{J}$.

Answer: $\Delta U=600\ \text{J}$.

6
Worked Example
Air ($\gamma=1.4$) at $27\ ^\circ\text{C}$ is compressed adiabatically to one-eighth of its volume. Find the final temperature.
Solution
  1. Step 1: For an adiabatic change $TV^{\gamma-1}=\text{const}$, so $T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
  2. Step 2: $T_1=300\ \text{K}$, $\frac{V_1}{V_2}=8$, $\gamma-1=0.4$: $T_2=300\times8^{0.4}$.
  3. Step 3: $8^{0.4}=2.297$, so $T_2=300\times2.297\approx689\ \text{K}$.

Answer: $T_2\approx689\ \text{K}\;(\approx416\ ^\circ\text{C})$.

Key Points

  • Isothermal ($T$ const): $\Delta U=0$, $\Delta Q=\Delta W$, $PV=$ const, $W=nRT\ln\!\left(\frac{V_2}{V_1}\right)$.
  • Adiabatic ($\Delta Q=0$): $\Delta W=-\Delta U$, $PV^\gamma=$ const, $TV^{\gamma-1}=$ const, $W=\frac{P_1V_1-P_2V_2}{\gamma-1}$.
  • Isobaric ($P$ const): $W=P\,\Delta V=nR\,\Delta T$, heat $\Delta Q=nC_p\,\Delta T$.
  • Isochoric ($V$ const): $W=0$, $\Delta Q=\Delta U=nC_v\,\Delta T$.
  • Mayer's relation $C_p-C_v=R$ and $\gamma=\frac{C_p}{C_v}$ ($\approx1.67$ monatomic, $1.40$ diatomic); adiabatic curve is steeper than the isotherm.
Quick Check — 4 MCQs
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Q1.In an isothermal process the change in internal energy of an ideal gas is:
Explanation: Internal energy depends only on temperature; at constant $T$, $\Delta U=0$.
Q2.An adiabatic process obeys the relation:
Explanation: With no heat exchange, $PV^\gamma=\text{const}$ where $\gamma=\frac{C_p}{C_v}$.
Q3.In which process is the work done by the gas zero?
Explanation: Isochoric means constant volume, so $dV=0$ and $W=\int P\,dV=0$.
Q4.Mayer's relation between the molar specific heats is:
Explanation: Applying the first law to one mole at constant pressure gives $C_p-C_v=R$.
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