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

Definition Symbols and sign conventions used everywhere on this page

Before any algebra, fix the meaning of every symbol so nothing is assumed:

$n$ = number of moles of gas — literally how many packets of $6 \times 1 0^{23}$ molecules you have. When a problem says " $2 mol$ ," that number is $n = 2$ ; drop it straight into every formula.
$R = 8.31 J mol^{- 1} K^{- 1}$ = the universal gas constant (the fixed conversion between moles-times-kelvin and joules).
$T$ = temperature (K), $P$ = pressure (Pa), $V$ = volume (m³).
$Q$ = heat. Sign convention: $Q > 0$ means heat is added into the gas; $Q < 0$ means heat leaves it.
$W$ = work. Sign convention: $W > 0$ means work is done by the gas (it expands, pushing out); $W < 0$ means work is done on the gas (it is compressed).
First Law (energy bookkeeping): $Δ U = Q - W$ . Internal energy rises by the heat you put in minus the work the gas hands out.
Ideal Gas Law: $P V = n R T$ links the three state variables through $n$ and $R$ .

Constants for gas types: Monatomic $C_{V} = \frac{3}{2} R$ , $C_{P} = \frac{5}{2} R$ , $γ = \frac{5}{3}$ . Diatomic $C_{V} = \frac{5}{2} R$ , $C_{P} = \frac{7}{2} R$ , $γ = \frac{7}{5}$ . Here $C_{V}, C_{P}$ are the heat capacities and $γ = C_{P} / C_{V}$ .

Level 1 — Recognition

These test which quantity is frozen and what that immediately forces to zero.

L1.1

On a $P$ – $V$ diagram a gas moves along a vertical line downward (pressure drops, volume unchanged). Name the process and state $W$ .

Recall Solution

WHAT the picture shows: volume never changes ⇒ this is the isochoric process. WHY $W = 0$ : work is the area under the curve, $W = \int P d V$ . If $V$ never changes then $d V = 0$ everywhere, so the "area" is a line with zero width. $W = \int P d V = 0.$ All energy change is heat becoming internal energy: $Q = Δ U = n C_{V} Δ T$ (recall $n$ = moles, defined above).

L1.2

A gas is compressed so fast (in an insulated cylinder) that no heat can flow. Which process is this, and what is $Q$ ?

Recall Solution

No heat exchanged ⇒ adiabatic, so $Q = 0$ . The First Law then reads $Δ U = Q - W = - W$ . Since we compress the gas, work is done on it, so by our sign convention $W < 0$ , giving $Δ U > 0$ : the gas heats up. (This is diesel ignition.)

L1.3

Match each curve to its process: (a) horizontal line, (b) hyperbola $P V =$ const, (c) vertical line, (d) steeper hyperbola $P V^{γ} =$ const.

Recall Solution

(a) horizontal → pressure fixed → isobaric.
(b) hyperbola $P V =$ const → temperature fixed (since $P V = n R T$ ) → isothermal.
(c) vertical → volume fixed → isochoric.
(d) steeper hyperbola → $Q = 0$ → adiabatic (steeper because $γ > 1$ ).

Level 2 — Application

Now plug numbers into one formula each.

L2.1

Seal $2 mol$ of monatomic gas in a rigid can and heat it from $300 K$ to $450 K$ . Find $W$ , $Δ U$ , and $Q$ .

Recall Solution

Here $n = 2$ (two moles). Rigid can ⇒ isochoric ⇒ $W = 0$ . $Δ U = n C_{V} Δ T = 2 \cdot \frac{3}{2} (8.31) (150) = 3739.5 J .$ First Law: $Q = Δ U + W = 3739.5 J$ . Every joule of heat became internal energy.

L2.2

$1 mol$ of gas expands isothermally at $350 K$ from $V_{1}$ to $V_{2} = 3 V_{1}$ . Find $W$ , $Δ U$ , $Q$ .

Recall Solution

Here $n = 1$ . Isothermal ⇒ $Δ U = 0$ (since $U$ depends only on $T$ ). $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.31) (350) ln 3 = 3195.4 J .$ $Q = W = 3195.4 J$ (all heat in comes straight out as work).

L2.3

$3 mol$ of gas expand isobarically at $P = 2.0 \times 1 0^{5} Pa$ from $V_{1} = 0.02 m^{3}$ to $V_{2} = 0.05 m^{3}$ . Find $W$ . Then, if the gas is diatomic, find $Q$ .

Recall Solution

Here $n = 3$ . Work (pressure constant ⇒ pulls out of the integral): $W = P Δ V = (2.0 \times 1 0^{5}) (0.05 - 0.02) = 6000 J .$ Temperature change — WHY $Δ T = P Δ V / (n R)$ : write the ideal gas law at the start, $P V_{1} = n R T_{1}$ , and at the end, $P V_{2} = n R T_{2}$ (same $P$ ). Subtract them: $P (V_{2} - V_{1}) = n R (T_{2} - T_{1})$ , i.e. $P Δ V = n R Δ T$ . Solve for $Δ T$ : $Δ T = \frac{P Δ V}{n R} = \frac{6000}{3 ( 8.31 )} = 240.67 K .$ Heat (diatomic $C_{P} = \frac{7}{2} R$ ): $Q = n C_{P} Δ T = 3 \cdot \frac{7}{2} (8.31) (240.67) = 21000 J .$ (Check: $Q = Δ U + W$ , and $Δ U = n C_{V} Δ T = 3 \cdot \frac{5}{2} (8.31) (240.67) = 15000 J$ ; $15000 + 6000 = 21000$ . ✓)

Level 3 — Analysis

Two-step reasoning, or the adiabat curve.

L3.1

$1 mol$ of monatomic gas ( $γ = \frac{5}{3}$ ) starts at $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.025 m^{3}$ and expands adiabatically to $V_{2} = 0.050 m^{3}$ . Find $P_{2}$ and the work done.

Recall Solution

WHAT/WHY: adiabatic ⇒ $Q = 0$ ⇒ curve is $P V^{γ} =$ const. Use it to find $P_{2}$ . $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = 1.0 \times 1 0^{5} (\frac{1}{2})^{5/3} = 3.150 \times 1 0^{4} Pa .$ Where the work formula comes from (WHY, not authority): since $Q = 0$ , the First Law says the gas can only do work by spending internal energy: $W = - Δ U = - n C_{V} (T_{2} - T_{1}) = n C_{V} (T_{1} - T_{2})$ . Now trade $n C_{V}$ for pressures and volumes. From $C_{V} = R / (γ - 1)$ we get $n C_{V} T = \frac{n R T}{γ - 1} = \frac{P V}{γ - 1}$ (using $P V = n R T$ ). Therefore $W = n C_{V} T_{1} - n C_{V} T_{2} = \frac{P _{1} V _{1}}{γ - 1} - \frac{P _{2} V _{2}}{γ - 1} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} .$ That is the "state formula" — it is just $- Δ U$ rewritten. Plug in: $W = \frac{( 1.0 \times 1 0 ^{5} ) ( 0.025 ) - ( 3.150 \times 1 0 ^{4} ) ( 0.050 )}{\frac{5}{3} - 1} .$ Numerator $= 2500 - 1575 = 925$ ; denominator $= \frac{2}{3}$ ; so $W = 1387.5 J$ . $W > 0$ (gas expands), and $Δ U = - W = - 1387.5 J$ — the gas cooled.

L3.2

Same start state as L3.1, but now expand isothermally to the same $V_{2} = 0.050 m^{3}$ . Find the final pressure $P_{2}^{'}$ and compare it to the adiabatic $P_{2}$ . Which curve is steeper? See the figure.

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

Recall Solution

Isotherm: $P V =$ const, so $P_{2}^{'} = P_{1} \frac{V _{1}}{V _{2}} = 1.0 \times 1 0^{5} \cdot \frac{1}{2} = 5.00 \times 1 0^{4} Pa$ . Compare: adiabatic gave $P_{2} = 3.15 \times 1 0^{4} Pa$ , which is lower. Why: both curves leave the same point, but the adiabat obeys $P V^{γ}$ with $γ = \frac{5}{3} > 1$ , so pressure falls faster. For the same expansion the adiabat lands below the isotherm — it is steeper. Reading the figure (in words, in case it doesn't load): the horizontal axis is volume $V$ in litres, the vertical axis is pressure $P$ in units of $1 0^{5} Pa$ . Both curves start together at the white dot $(25 L, 1.0)$ and slope downward to the right. The cyan curve is the gentle isotherm $P V =$ const, ending near $P = 0.5$ . The amber curve is the steeper adiabat $P V^{γ} =$ const, which drops faster and ends lower, near $P = 0.315$ — visibly underneath the cyan curve at the same volume. That vertical gap between them at $V = 50 L$ is the whole point: the adiabat loses more pressure because it gets no heat top-up.

L3.3

A gas absorbs $Q = 800 J$ while its internal energy rises by $Δ U = 500 J$ . Find the work done by the gas, and state whether it expanded or was compressed.

Recall Solution

First Law: $W = Q - Δ U = 800 - 500 = 300 J$ . $W > 0$ ⇒ work done by the gas ⇒ it expanded.

Level 4 — Synthesis

Combine legs into a whole cycle.

L4.1

$1 mol$ of monatomic gas runs a 3-leg cycle (see figure):

A→B isochoric heating: $P$ rises from $1.0 \times 1 0^{5}$ to $2.0 \times 1 0^{5} Pa$ at $V = 0.020 m^{3}$ .
B→C isobaric expansion: $V$ rises from $0.020$ to $0.040 m^{3}$ at $P = 2.0 \times 1 0^{5} Pa$ .
C→A straight line back to A.

Find $W$ for each leg and the net work of the cycle (= enclosed area).

Recall Solution

A→B isochoric: $d V = 0 \Rightarrow W_{A B} = 0$ . B→C isobaric: $W_{B C} = P Δ V = (2.0 \times 1 0^{5}) (0.040 - 0.020) = 4000 J$ . C→A straight line from $(V = 0.040, P = 2.0 \times 1 0^{5})$ to $(V = 0.020, P = 1.0 \times 1 0^{5})$ . Work = area of a trapezoid under this line (volume decreases, so $W_{C A} < 0$ ): $W_{C A} = - trapezoid \frac{1}{2} (P_{C} + P_{A}) (V_{C} - V_{A}) = - \frac{1}{2} (2.0 \times 1 0^{5} + 1.0 \times 1 0^{5}) (0.020) = - 3000 J .$ Net work = area enclosed by the triangle A-B-C: $W_{net} = 0 + 4000 - 3000 = 1000 J .$ Cross-check with the triangle area: base $= Δ V = 0.020$ , height $= Δ P = 1.0 \times 1 0^{5}$ , area $= \frac{1}{2} (0.020) (1.0 \times 1 0^{5}) = 1000 J$ . ✓ Positive ⇒ clockwise cycle ⇒ it acts as an engine. Reading the figure (in words): the axes are volume $V$ (litres) horizontally and pressure $P$ (units of $1 0^{5} Pa$ ) vertically. Point A sits low-left at $(20, 1.0)$ , B directly above it at $(20, 2.0)$ , and C to the right at $(40, 2.0)$ . The cyan vertical segment A→B is the isochoric leg (no width ⇒ no work). The amber horizontal segment B→C is the isobaric expansion (the rectangle under it is the $+ 4000 J$ ). The white slanted segment C→A closes the loop. The lightly shaded triangle inside the loop is the net work $1000 J$ — literally the area enclosed. The loop runs clockwise, the signature of an engine.

L4.2

For the same cycle, find $Δ U$ over the whole cycle, and hence the net heat $Q_{net}$ .

Recall Solution

WHY $Δ U_{cycle} = 0$ : internal energy $U$ is a state function — it depends only on where the gas is (its $P, V, T$ ), not on the path taken. A cycle ends at the very same state it began, so $T$ returns to its starting value, giving $Δ T = 0$ . Then $Δ U_{cycle} = n C_{V} Δ T = n C_{V} (0) = 0$ . First Law over the cycle: $Q_{net} = Δ U_{cycle} + W_{net} = 0 + 1000 = 1000 J$ . Net heat in equals net work out — that is the engine's job.

Level 5 — Mastery

Real algebra, subtle traps.

L5.1

$2 mol$ of diatomic gas ( $γ = \frac{7}{5}$ ) is compressed adiabatically from $T_{1} = 300 K$ , $V_{1}$ to $V_{2} = V_{1} /4$ . Find $T_{2}$ and the work done on the gas.

Recall Solution

Here $n = 2$ . WHY use $T V^{γ - 1}$ : we know volumes and a temperature, not pressures — so pick the $T$ – $V$ form of the adiabatic law. $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \cdot 4^{2/5} = 300 \cdot 1.7411 = 522.3 K .$ Work done by gas (same $- Δ U$ reasoning as L3.1): $W = n C_{V} (T_{1} - T_{2}) = 2 \cdot \frac{5}{2} (8.31) (300 - 522.3) = - 9239 J$ . Negative ⇒ work is done on the gas. Work done on the gas $= + 9239 J$ . (Compression heats it: $T$ jumped from $300$ to $522 K$ .)

L5.2

A monatomic gas expands, doing $W = 1200 J$ , while $600 J$ of heat is added. Was this expansion isothermal, adiabatic, or neither? Justify.

Recall Solution

Compute $Δ U = Q - W = 600 - 1200 = - 600 J$ .

Isothermal would need $Δ U = 0$ — not the case.
Adiabatic would need $Q = 0$ — but $Q = 600 J \neq = 0$ . So it is neither. The internal energy dropped ( $Δ U < 0$ , gas cooled) even though heat was added, because the gas did more work than the heat supplied — the extra came from internal energy.

L5.3

Starting from $(P_{1}, V_{1})$ , a monatomic gas can reach volume $2 V_{1}$ two ways: (a) isothermally, (b) adiabatically. In which case does the gas do more work, and why? Prove it with the two work formulas for the numbers $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.030 m^{3}$ , $T_{1}$ from $n R T_{1} = P_{1} V_{1}$ .

Recall Solution

Isothermal work: $W_{iso} = P_{1} V_{1} ln \frac{V _{2}}{V _{1}} = (1.0 \times 1 0^{5}) (0.030) ln 2 = 2079.4 J$ . (Used $n R T_{1} = P_{1} V_{1} = 3000 J$ .) Adiabatic work (state formula from L3.1, which was itself $- Δ U$ ): $P_{2} = P_{1} (V_{1} / V_{2})^{γ} = 1.0 \times 1 0^{5} \cdot (1/2)^{5/3} = 3.150 \times 1 0^{4} Pa$ . $W_{adi} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{3000 - ( 3.150 \times 1 0 ^{4} ) ( 0.060 )}{2/3} = \frac{3000 - 1890}{0.6667} = 1665 J .$ Isothermal does more work ( $2079 > 1665 J$ ). Why: in the isothermal path heat is continuously fed in to keep pressure up along the way, so the gas pushes harder at every volume. In the adiabatic path there is no heat top-up, so pressure sags below the isotherm — less force, less area, less work. (This is exactly the vertical gap you saw in the L3.2 figure.)

Self-test recall

Which quantity is a state function, so its change is zero around any cycle?

Internal energy

U

(thus

Δ U_{cycle} = 0

, giving

Q_{net} = W_{net}

Which adiabatic form do you use when given

T

and

V

T V^{γ - 1} = const

For the same expansion, which does more work — isothermal or adiabatic?

Isothermal (heat top-up keeps pressure higher along the path).

Where does the adiabatic state work formula

W = (P_{1} V_{1} - P_{2} V_{2}) / (γ - 1)

come from?

It is just

W = - Δ U = n C_{V} (T_{1} - T_{2})

rewritten using

n C_{V} T = P V / (γ - 1)

Related: Carnot Cycle builds an engine from two isotherms + two adiabats; Entropy measures the heat/temperature bookkeeping across these legs.