Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0): Worked examples

Throughout, $R = 8.31 J mol^{- 1} K^{- 1}$ . Monatomic gas: $C_{V} = \frac{3}{2} R$ , $C_{P} = \frac{5}{2} R$ , $γ = \frac{5}{3}$ . Diatomic gas: $C_{V} = \frac{5}{2} R$ , $C_{P} = \frac{7}{2} R$ , $γ = \frac{7}{5}$ .

The scenario matrix

Every problem in this chapter is one of these cells. Each column is a "sign or degenerate axis"; every worked example below is tagged with the cell it covers.

Cell	Description	Sign / limiting feature	Example
A	Isochoric heating	$W = 0$ , $Q = Δ U > 0$	Ex 1
B	Isobaric expansion	$W > 0$ , split $Q = Δ U + W$	Ex 2
C	Isothermal expansion	$Δ U = 0$ , $Q = W > 0$	Ex 3
D	Isothermal compression	$Δ U = 0$ , $Q = W < 0$ (sign flip)	Ex 3b
E	Adiabatic expansion	$Q = 0$ , gas cools, $W > 0$	Ex 4
F	Adiabatic compression	$Q = 0$ , gas heats, $W < 0$ (sign flip)	Ex 5
G	Compare isotherm vs adiabat from same start	steepness / $γ > 1$ limiting geometry	Ex 6
H	Real-world word problem	atmosphere / diesel	Ex 5 (diesel)
I	Exam twist: two-leg cycle, net $W$	path dependence of $W$ , $Δ U = 0$ over cycle	Ex 7
J	Degenerate limit: $V_{2} \to V_{1}$	$ln 1 = 0$ , all quantities $\to 0$	Ex 8

We now walk every cell.

Example 1 — Cell A: Isochoric heating

Example 2 — Cell B: Isobaric expansion (the energy split)

Worked example 2 mol monatomic gas at constant

$P = 1.0 \times 1 0^{5} Pa$ , heated so volume goes $V_{1} = 0.020 m^{3} \to V_{2} = 0.035 m^{3}$ . Find $W$ , $Δ T$ , $Δ U$ , $Q$ , and check that $Q$ splits into $Δ U + W$ .

Forecast: the gas both heats and pushes the piston. Guess: will $Q$ be bigger or smaller than $Δ U$ ?

Step 1. $W = P Δ V = (1.0 \times 1 0^{5}) (0.035 - 0.020) = 1500 J$ . Why this step? $P$ is constant so it pulls out of $\int P d V$ , leaving $P Δ V$ — a rectangle area under the horizontal line (see Work done by gas — PV diagrams).

Step 2. From $P Δ V = n R Δ T$ : $Δ T = \frac{P Δ V}{n R} = \frac{1500}{2 \cdot 8.31} = 90.25 K$ . Why this step? $P V = n R T$ at constant $P$ gives $P Δ V = n R Δ T$ ; we need $Δ T$ to get $Δ U$ and $Q$ .

Step 3. $Δ U = n C_{V} Δ T = 2 \cdot \frac{3}{2} (8.31) (90.25) = 2250 J$ .

Step 4. $Q = n C_{P} Δ T = 2 \cdot \frac{5}{2} (8.31) (90.25) = 3750 J$ . Why this step? Along an isobar, heat is measured by $C_{P}$ (from Heat Capacities Cv and Cp), which already bundles the expansion cost.

Verify: $Δ U + W = 2250 + 1500 = 3750 = Q$ ✓. Confirms Mayer's split: $C_{P} > C_{V}$ because the extra $R$ per mole paid the $1500 J$ of work. $Q > Δ U$ , matching the forecast.

Example 3 — Cell C: Isothermal expansion

Worked example 1 mol gas at

$T = 320 K$ , volume expands from $V_{1}$ to $V_{2} = 3 V_{1}$ in contact with a reservoir. Find $Δ U$ , $W$ , $Q$ .

Forecast: temperature is pinned by the reservoir. Guess: what is $Δ U$ ? Where does the work energy come from?

Step 1. $T$ constant $\Rightarrow Δ U = n C_{V} Δ T = 0$ . Why this step? $U$ depends only on $T$ ; no temperature change means no internal-energy change.

Step 2. First Law: $0 = Q - W \Rightarrow Q = W$ . Why this step? With $Δ U = 0$ , every joule of heat in comes straight out as work.

Step 3. $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.31) (320) ln 3 = 2921.6 J$ . Why this step? $P = n R T / V$ ; $n R T$ is constant so it leaves the integral, and $\int d V / V = ln V$ . We use $ln$ because the integrand $1/ V$ is exactly the derivative of $ln V$ — no other function answers "what has slope $1/ V$ ?"

Verify: $Q = W = 2921.6 J$ , $Δ U = 0$ ✓. Positive $W$ ⇒ expansion ⇒ heat absorbed, correct for an isothermal expansion.

Example 3b — Cell D: Isothermal compression (sign flip)

Example 4 — Cell E: Adiabatic expansion (gas cools)

Worked example 1 mol monatomic gas (

$γ = \frac{5}{3}$ ) expands adiabatically. Start: $P_{1} = 2.0 \times 1 0^{5} Pa$ , $V_{1} = 0.010 m^{3}$ . It expands to $V_{2} = 0.020 m^{3}$ . Find $P_{2}$ , $T_{1}$ , $T_{2}$ , and $W$ .

Forecast: no heat can enter. Guess: as it expands, does the gas heat up or cool down? Where does the work come from?

Step 1. $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = (2.0 \times 1 0^{5}) (\frac{1}{2})^{5/3} = 63000 Pa$ (to 2 s.f. $6.3 \times 1 0^{4}$ ). Why this step? Adiabats obey $P V^{γ} = const$ , so $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ .

Step 2. $T_{1} = \frac{P _{1} V _{1}}{n R} = \frac{( 2.0 \times 1 0 ^{5} ) ( 0.010 )}{( 1 ) ( 8.31 )} = 240.7 K$ .

Step 3. $T_{2} = \frac{P _{2} V _{2}}{n R} = \frac{( 63000 ) ( 0.020 )}{8.31} = 151.6 K$ . Why this step? Once $P_{2}$ and $V_{2}$ are known, $P V = n R T$ gives the new temperature directly. $T_{2} < T_{1}$ : the gas cooled, exactly the forecast.

Step 4. $W = n C_{V} (T_{1} - T_{2}) = (1) \frac{3}{2} (8.31) (240.7 - 151.6) = 1111 J$ . Why this step? $Q = 0$ ⇒ $W = - Δ U = n C_{V} (T_{1} - T_{2})$ ; the work is paid entirely out of internal energy.

Verify: Cross-check with $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{2000 - 1260}{2/3} = 1110 J$ ✓ (rounding). $W > 0$ and $T$ dropped — consistent with an insulated expansion (this is how clouds form as air rises).

Example 5 — Cell F & H: Adiabatic compression (diesel ignition)

Worked example Air (

$γ = 1.4$ , treat as diesel) in a cylinder is compressed adiabatically from $V_{1}$ to $V_{2} = \frac{1}{15} V_{1}$ (compression ratio 15). Initial temperature $T_{1} = 300 K$ . Find $T_{2}$ and the sign of $W$ .

Forecast: we squeeze fast, no heat escapes. Guess: does the temperature rise enough (~600 K) to ignite diesel?

Step 1. $T V^{γ - 1} = const \Rightarrow T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \cdot 1 5^{0.4}$ . Why this step? The $T$ – $V$ form of the adiabat links temperatures directly to the volume ratio, skipping pressure.

Step 2. $1 5^{0.4} = 2.954$ , so $T_{2} = 300 \times 2.954 = 886.3 K$ . Why this step? $886 K$ (≈ 613 °C) is well above diesel's autoignition (~483 K) — the fuel ignites with no spark plug. That is the entire principle of a diesel engine.

Step 3. Sign of $W$ : $T_{2} > T_{1}$ so $Δ U > 0$ , and $W = - Δ U < 0$ . Why this step? Work is done on the gas during compression, so $W$ (work by gas) is negative — the classic sign trap from the parent note.

Verify: $T_{2} = 886 K > T_{1}$ ✓ heating on compression, $W < 0$ ✓. Real-world (Cell H) satisfied: adiabatic compression is what ignites diesel.

Example 6 — Cell G: Isotherm vs adiabat from the same start

Worked example Both an isotherm and an adiabat (

$γ = \frac{5}{3}$ ) start at $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.010 m^{3}$ and expand to $V_{2} = 0.020 m^{3}$ . Find the final pressure on each curve and confirm which drops lower.

Forecast: same start, same expansion. Guess: which curve ends at the lower pressure?

Step 1. Isotherm ( $P V = const$ ): $P_{2}^{iso} = P_{1} \frac{V _{1}}{V _{2}} = (1.0 \times 1 0^{5}) (0.5) = 50000 Pa$ . Why this step? On an isotherm pressure falls in simple inverse proportion to volume.

Step 2. Adiabat ( $P V^{γ} = const$ ): $P_{2}^{ad} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = (1.0 \times 1 0^{5}) (0.5)^{5/3} = 31498 Pa$ . Why this step? The exponent $γ > 1$ makes the same volume ratio produce a larger pressure drop.

Step 3. Compare: $31498 < 50000$ . The adiabat ends lower — it is steeper. Why this step? On the adiabat the gas also loses temperature (no heat replaces the energy spent on work), so pressure sags further. Look at the figure: at $V_{2}$ the violet adiabat sits below the magenta isotherm.

Verify: ratio $\frac{P _{2}^{ad}}{P _{2}^{iso}} = \frac{31498}{50000} = 0.63 = (0.5)^{γ - 1} = (0.5)^{2/3}$ ✓. The steepness factor is exactly $(V_{1} / V_{2})^{γ - 1}$ , the limiting-geometry signature of $γ > 1$ .

Example 7 — Cell I: Two-leg cycle, net work (exam twist)

Worked example 1 mol monatomic gas runs:

Leg 1 isobaric expansion at $P = 1.0 \times 1 0^{5} Pa$ from $V_{A} = 0.010$ to $V_{B} = 0.020 m^{3}$ ; Leg 2 isochoric cooling at $V = 0.020 m^{3}$ back to the pressure such that a final isothermal leg would return it... — simplify: after Leg 1 and Leg 2, the gas is at $P_{C} = 0.5 \times 1 0^{5} Pa$ , $V_{C} = 0.020 m^{3}$ . Find total work over the two legs.

Forecast: work depends on the path. Guess: does the isochoric leg contribute to work?

Step 1. Leg 1 (isobaric): $W_{1} = P Δ V = (1.0 \times 1 0^{5}) (0.020 - 0.010) = 1000 J$ . Why this step? Rectangle area under the horizontal line.

Step 2. Leg 2 (isochoric): $W_{2} = 0$ . Why this step? $d V = 0$ , no swept area — the cornerstone of Cell A applies mid-cycle too.

Step 3. Total $W = W_{1} + W_{2} = 1000 + 0 = 1000 J$ . Why this step? Work over a multi-leg path is the sum of leg works — $W$ is path-dependent, unlike $Δ U$ .

Verify: only the expanding leg does work; $1000 J$ ✓. If this were a closed cycle, $\oint d U = 0$ so $Q_{net} = W_{net}$ — the seed of the Carnot Cycle and Entropy bookkeeping.

Example 8 — Cell J: Degenerate limit $V_{2} \to V_{1}$

Recall Self-test: name the cell

A gas is compressed with no heat exchange and its temperature rises. Which cell? ::: Cell F (adiabatic compression, $W < 0$ ). A gas expands at fixed temperature; what is $Δ U$ ? ::: $0$ (Cell C/D). Why does the adiabat drop below the isotherm at the same $V_{2}$ ? ::: Exponent $γ > 1$ gives a larger pressure drop; the gas also cools (Cell G). In a sealed rigid can heated up, what is $W$ ? ::: $0$ (Cell A). Which quantity is path-dependent across a multi-leg cycle? ::: Work $W$ (and $Q$ ); $Δ U$ is not (Cell I).

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