Work done in each process — derivation: Exercises

Constants used throughout: gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ , and $ln 2 = 0.6931$ , $ln 3 = 1.0986$ .

Level 1 — Recognition

Goal: pick the right formula and plug in. No traps in the algebra, only in choosing.

The four tools you are choosing between:

Q1.1 (L1). A gas expands at a constant pressure of $2.0 \times 1 0^{5} Pa$ from $V_{1} = 1.0 \times 1 0^{- 3} m^{3}$ to $V_{2} = 3.0 \times 1 0^{- 3} m^{3}$ . Find the work done by the gas.

Recall Solution

Which tool? Pressure is constant → isobaric → $W = P (V_{2} - V_{1})$ . $W = 2.0 \times 1 0^{5} \times (3.0 \times 1 0^{- 3} - 1.0 \times 1 0^{- 3}) = 2.0 \times 1 0^{5} \times 2.0 \times 1 0^{- 3}$ $W = 400 J .$ What it looks like: on a P–V diagram this is a flat horizontal line; the area under it is a rectangle of height $P$ and width $Δ V$ . Positive because the gas expanded ( $V_{2} > V_{1}$ ).

Q1.2 (L1). A gas is heated inside a rigid sealed steel tank from $300 K$ to $500 K$ . Find the work done by the gas.

Recall Solution

Which tool? "Rigid tank" = volume cannot change → isochoric → $d V = 0$ . $W = \int_{V_{1}}^{V_{1}} P d V = 0 J .$ The piston (walls) never moves, so no distance is pushed through — no work, no matter how much the temperature or pressure climbs.

Q1.3 (L1). $1$ mol of ideal gas expands isothermally at $T = 400 K$ until its volume triples. Find $W$ .

Recall Solution

Which tool? Temperature constant → isothermal → $W = n R T ln (V_{2} / V_{1})$ with $V_{2} / V_{1} = 3$ . $W = 1 \times 8.314 \times 400 \times ln 3 = 3325.6 \times 1.0986 = 3653.4 J .$ Positive: expansion.

Level 2 — Application

Goal: convert units, pull out $n R T$ , use $P V = n R T$ before you can plug in.

Q2.1 (L2). $2$ mol of gas is heated at constant pressure from $300 K$ to $450 K$ . Find $W$ using the temperature form.

Recall Solution

Isobaric, so $W = n R Δ T$ (this is the same rectangle as $P Δ V$ , rewritten with the ideal gas law $P V = n R T$ ). $W = 2 \times 8.314 \times (450 - 300) = 2 \times 8.314 \times 150 = 2494.2 J .$

Q2.2 (L2). $0.5$ mol of gas expands isothermally at $350 K$ from $4.0 L$ to $10.0 L$ . Find $W$ .

Recall Solution

Which tool? Isothermal → $W = n R T ln (V_{2} / V_{1})$ . The litres cancel inside the ratio, so no unit conversion is needed there. $\frac{V _{2}}{V _{1}} = \frac{10.0}{4.0} = 2.5, ln 2.5 = 0.9163.$ $W = 0.5 \times 8.314 \times 350 \times 0.9163 = 1454.95 \times 0.9163 = 1333.2 J .$

Q2.3 (L2). $1$ mol of a monatomic gas ( $γ = 5/3$ ) expands adiabatically. Its temperature falls from $500 K$ to $350 K$ . Find $W$ .

Recall Solution

Which tool? $Q = 0$ → adiabatic → $W = \frac{n R ( T _{1} - T _{2} )}{γ - 1}$ . $γ - 1 = \frac{5}{3} - 1 = \frac{2}{3} = 0.6667.$ $W = \frac{1 \times 8.314 \times ( 500 - 350 )}{0.6667} = \frac{1247.1}{0.6667} = 1870.7 J .$ Positive because it expanded ( $T_{1} > T_{2}$ ). The adiabatic work came entirely out of the gas's own internal energy, which is why it cooled.

Level 3 — Analysis

Goal: combine with the first law, read signs, work backwards.

Q3.1 (L3). During an isothermal compression, $1$ mol of gas at $300 K$ is squeezed from $6.0 L$ to $2.0 L$ . (a) Find the work done by the gas. (b) Using the First Law of Thermodynamics, find the heat $Q$ .

Recall Solution

(a) Isothermal → $W = n R T ln (V_{2} / V_{1})$ with $V_{2} / V_{1} = 2/6 = 1/3$ . $W = 1 \times 8.314 \times 300 \times ln (\frac{1}{3}) = 2494.2 \times (- 1.0986) = - 2740.1 J .$ Negative — the gas is compressed, so work by the gas is negative (work is done on it). (b) Isothermal $\Rightarrow Δ T = 0 \Rightarrow Δ U = 0$ . First law: $Δ U = Q - W$ , so $Q = W = - 2740.1 J$ . $Q < 0$ : heat must leave the gas to keep $T$ fixed while it is being compressed.

Q3.2 (L3). A gas does $500 J$ of work in an adiabatic expansion. What is the change in its internal energy $Δ U$ ? Does it warm or cool?

Recall Solution

Adiabatic means $Q = 0$ . First law: $Δ U = Q - W = 0 - 500 = - 500 J$ . Internal energy dropped by $500 J$ , so the gas cools. All the work was paid for out of internal energy — no heat came in to refill it.

Q3.3 (L3). Look at the figure below. A gas goes from state $A$ $(P = 3 \times 1 0^{5} Pa, V = 2 L)$ to state $B$ $(P = 1 \times 1 0^{5} Pa, V = 6 L)$ along a straight line on the P–V diagram. Find the work done.

Figure — Work done in each process — derivation

Recall Solution

Idea: $W$ = area under the curve. The path is a straight line, so the area under it is a trapezium (a rectangle plus a triangle), bounded below by the $V$ -axis. Convert litres to $m^{3}$ : $2 L = 2 \times 1 0^{- 3} m^{3}$ , $6 L = 6 \times 1 0^{- 3} m^{3}$ . Trapezium area = average height × width: $W = \frac{P _{A} + P _{B}}{2} (V_{B} - V_{A}) = \frac{( 3 + 1 ) \times 1 0 ^{5}}{2} \times (6 - 2) \times 1 0^{- 3}$ $W = 2 \times 1 0^{5} \times 4 \times 1 0^{- 3} = 800 J .$ Positive: the path moves rightward ( $V$ increases). Look at the shaded region in the figure — that is the $800 J$ .

Level 4 — Synthesis

Goal: chain several processes; add works with correct signs.

Q4.1 (L4). A gas is taken through three steps:

Isobaric expansion at $P = 2 \times 1 0^{5} Pa$ from $1 L$ to $3 L$ .
Isochoric cooling at $V = 3 L$ (pressure drops to $1 \times 1 0^{5} Pa$ ).
Isobaric compression at $P = 1 \times 1 0^{5} Pa$ back from $3 L$ to $1 L$ .

Find the total work done by the gas over the three steps.

Recall Solution

Do each step with its own tool, then add. Step 1 (isobaric expand): $W_{1} = P Δ V = 2 \times 1 0^{5} \times (3 - 1) \times 1 0^{- 3} = + 400 J$ . Step 2 (isochoric): $d V = 0 \Rightarrow W_{2} = 0 J$ . Step 3 (isobaric compress): $W_{3} = P Δ V = 1 \times 1 0^{5} \times (1 - 3) \times 1 0^{- 3} = - 200 J$ (negative — volume shrinks). $W_{total} = 400 + 0 - 200 = + 200 J .$ The steps do not close the loop yet (pressure at the end is $1 \times 1 0^{5}$ , start was $2 \times 1 0^{5}$ ), so this is an open path, not a cycle.

Q4.2 (L4). A complete rectangular cycle on a P–V diagram (see figure) runs clockwise: $A \to B \to C \to D \to A$ with corners $A (1 \times 1 0^{5} Pa, 2 L)$ , $B (3 \times 1 0^{5} Pa, 2 L)$ , $C (3 \times 1 0^{5} Pa, 5 L)$ , $D (1 \times 1 0^{5} Pa, 5 L)$ . Find the net work per cycle.

Recall Solution

Idea: for a closed loop, $W_{net} = \oint P d V$ = area enclosed. A clockwise loop gives positive net work. The rectangle has:

height $Δ P = (3 - 1) \times 1 0^{5} = 2 \times 1 0^{5} Pa$ ,
width $Δ V = (5 - 2) \times 1 0^{- 3} = 3 \times 1 0^{- 3} m^{3}$ . $W_{net} = Δ P \times Δ V = 2 \times 1 0^{5} \times 3 \times 1 0^{- 3} = 600 J .$ Check by summing sides (only the two horizontal legs move volume):
$B \to C$ (isobaric expand at $3 \times 1 0^{5}$ ): $+ 3 \times 1 0^{5} \times 3 \times 1 0^{- 3} = + 900 J$
$D \to A$ (isobaric compress at $1 \times 1 0^{5}$ ): $- 1 \times 1 0^{5} \times 3 \times 1 0^{- 3} = - 300 J$
vertical legs $A \to B$ , $C \to D$ : $Δ V = 0 \Rightarrow 0 J$ Total $= 900 - 300 = 600 J$ . ✓ Matches the enclosed area.

Level 5 — Mastery

Goal: full derivations, mixed processes, work backwards from given work.

Q5.1 (L5). $1$ mol of a diatomic gas ( $γ = 7/5$ ) starts at $P_{1} = 4 \times 1 0^{5} Pa$ , $V_{1} = 2 \times 1 0^{- 3} m^{3}$ and expands adiabatically to $V_{2} = 8 \times 1 0^{- 3} m^{3}$ . Find (a) the final pressure $P_{2}$ , and (b) the work done.

Recall Solution

(a) Adiabatic law: $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \Rightarrow P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ}$ . $\frac{V _{1}}{V _{2}} = \frac{2}{8} = \frac{1}{4}, γ = 1.4.$ $(\frac{1}{4})^{1.4} = 4^{- 1.4} = e^{- 1.4 l n 4} = e^{- 1.4 \times 1.3863} = e^{- 1.9409} = 0.14359.$ $P_{2} = 4 \times 1 0^{5} \times 0.14359 = 5.744 \times 1 0^{4} Pa .$ (b) Adiabatic work $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ . $P_{1} V_{1} = 4 \times 1 0^{5} \times 2 \times 1 0^{- 3} = 800 J .$ $P_{2} V_{2} = 5.744 \times 1 0^{4} \times 8 \times 1 0^{- 3} = 459.5 J .$ $W = \frac{800 - 459.5}{1.4 - 1} = \frac{340.5}{0.4} = 851.3 J .$ Positive: expansion. The gas cooled because this work drained its internal energy.

Q5.2 (L5). An isothermal expansion of $2$ mol at $T = 320 K$ does $W = 3000 J$ of work. Find the ratio $V_{2} / V_{1}$ .

Recall Solution

Work backwards from $W = n R T ln (V_{2} / V_{1})$ . Solve for the log first. $ln \frac{V _{2}}{V _{1}} = \frac{W}{n R T} = \frac{3000}{2 \times 8.314 \times 320} = \frac{3000}{5320.96} = 0.56380.$ Exponentiate to undo the $ln$ : $\frac{V _{2}}{V _{1}} = e^{0.56380} = 1.757.$ So the gas expands to about $1.76 \times$ its starting volume.

Q5.3 (L5). Derive the isothermal work formula from scratch and then evaluate it for $1$ mol at $300 K$ expanding from $V_{1}$ to $V_{2} = 2 V_{1}$ . Confirm you get the same number as the parent-note example.

Recall Solution

Derivation (WHAT / WHY / WHAT IT LOOKS LIKE):

Start from the master formula $W = \int_{V_{1}}^{V_{2}} P d V$ — because work is the area under the P–V curve.
$P$ is not constant, so substitute the ideal gas law $P V = n R T \Rightarrow P = \frac{n R T}{V}$ — this turns $P$ into a known function of $V$ so we can actually integrate.
Since $T$ is constant, $n R T$ is a fixed number; pull it out: $W = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} .$
The integral $\int \frac{d V}{V} = ln V$ — this is the one function whose slope is $1/ V$ , which is exactly why a $1/ V$ curve gives a log-shaped area. $W = n R T [ln V]_{V_{1}}^{V_{2}} = n R T ln \frac{V _{2}}{V _{1}} .$ Evaluate with $V_{2} / V_{1} = 2$ : $W = 1 \times 8.314 \times 300 \times ln 2 = 2494.2 \times 0.6931 = 1728.8 J .$ This matches the parent note's isothermal example exactly. ✓

Recall Self-test summary (cover the right side)

Tool for constant pressure ::: $W = P (V_{2} - V_{1}) = n R Δ T$ Tool for constant volume ::: $W = 0$ Tool for constant temperature ::: $W = n R T ln (V_{2} / V_{1})$ Tool for $Q = 0$ (adiabatic) ::: $W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ Net work in a cycle ::: enclosed area; clockwise positive To recover $V_{2} / V_{1}$ from isothermal $W$ ::: $V_{2} / V_{1} = e^{W / (n R T)}$

Connections

Parent: Work done in each process — the master formula these exercises drill
First Law of Thermodynamics — used in Q3.1, Q3.2 to get $Q$ and $Δ U$
P-V Diagrams — Q3.3, Q4.2 are pure "area under / inside the curve"
Adiabatic Process and PV^gamma — Q2.3, Q5.1
Isothermal Process and Boyle's Law — Q3.1, Q5.2, Q5.3
Internal Energy and Cv — why adiabatic work cools the gas
Carnot Cycle — the cyclic-area idea of Q4.2, taken further