First law of thermodynamics

Mid-October 2026: in Year 4, at technical institutes of the technology track (and in many licei, the Italian academic high schools), in class you are still between heat and the gas laws. The first law of thermodynamics usually arrives in November, and the test that puts it to the proof falls five to nine weeks later, between the end of November and mid-December. I am writing now because this topic rests on a base that has to be built calmly: the equation of state, the $p$ - $V$ plane and, above all, a single sign convention that you must never betray.

In my lessons the first law is almost always lost for the same reason: the student knows the formula, but can’t say whether in that transformation the work is positive or negative, and then one wrong sign drags everything else down. The rest is mechanics: four transformations, four rules. If you lack the basics on gases, first review the gas laws (there I see them from the chemistry side) and specific heat and thermal equilibrium.

System, internal energy, and what the law says

A thermodynamic system is the portion of the world you are studying: here it will almost always be a gas in a cylinder with a piston. Everything else is the environment. The system has an internal energy $U$ , the sum of the kinetic and potential energy of its molecules. For an ideal gas $U$ depends only on temperature: if $T$ doesn’t change, $U$ doesn’t change.

The system can exchange energy with the environment in two ways: as heat $Q$ (because of a temperature difference) and as work $L$ (the piston moving). The first law is the conservation of energy written for these two exchanges:

Δ U = Q - L

The sign convention, written once and for all

In this post I use the convention most common in Italian schools:

$Q > 0$ : the system absorbs heat; $Q < 0$ : it gives it off.
$L > 0$ : the work is done by the system on the environment (the gas expands); $L < 0$ : it is done on the system (the gas is compressed).

Some books instead use $W$ for the work done on the gas and write $Δ U = Q + W$ . It is the same physics with the sign of $L$ reversed. Check in your book and notes which one your teacher uses, and then don’t change halfway through a test. Whoever mixes the two notations gets it wrong.

Equation of state and molar specific heats

For an ideal gas:

p V = n R T

with $p$ in pascals, $V$ in cubic metres, $T$ in kelvin, $n$ in moles and $R \approx 8.314 J/(mol K)$ . Remember the conversions: $1 L = 1 0^{- 3} m^{3}$ , $1 atm \approx 1.013 \cdot 1 0^{5} Pa$ , $T (K) = t (^{\circ} C) + 273.15$ . Here the temperature must be absolute, unlike the specific heat formulas where the difference is enough.

For a monatomic ideal gas (helium, neon, argon) the molar specific heats are

c_{V} = \frac{3}{2} R c_{P} = \frac{5}{2} R

The first applies to heating at constant volume, the second at constant pressure, and $c_{P} - c_{V} = R$ . The useful fact is that for an ideal gas, in any transformation,

Δ U = n c_{V} Δ T

even when the transformation is not at constant volume: $U$ depends only on $T$ , and $c_{V}$ is its coefficient. Almost all exercises are solved by calculating $Δ U$ from here, the work from the graph or from the formula, and the heat by difference.

Work is an area

On the $p$ - $V$ plane the work done by the gas in a transformation is the area under the curve:

expansion (the point moves to the right): $L > 0$ ;
compression (to the left): $L < 0$ ;
at constant volume the point doesn’t move horizontally: $L = 0$ ;
in a cycle (you return to the starting point) the total work is the enclosed area, positive if the cycle runs clockwise, negative if anticlockwise.

If the pressure is constant the calculation is a rectangle: $L = p Δ V$ . If it changes, you need the formula for the specific transformation.

The four transformations

Transformation	What stays constant	Work $L$	$Δ U$	Heat $Q$
isochoric	$V$	$0$	$n c_{V} Δ T$	$= Δ U$
isobaric	$p$	$p Δ V = n R Δ T$	$n c_{V} Δ T$	$n c_{P} Δ T$
isothermal	$T$	$n R T ln (V_{2} / V_{1})$	$0$	$= L$
adiabatic	no heat exchange	$= - Δ U$	$n c_{V} Δ T$	$0$

What the first law says in each case:

Isochoric: the piston is locked, $L = 0$ , so $Δ U = Q$ . All the heat ends up as internal energy and the temperature changes.
Isobaric: $Q = Δ U + L$ . The heat is split: part heats the gas, part moves the piston. That is why $c_{P} > c_{V}$ : to rise one kelvin at constant pressure you need more heat, because a fraction goes into work.
Isothermal: $T$ constant, $Δ U = 0$ , so $Q = L$ . A gas that expands absorbs heat and gives it all back as work. The fact that the formula contains a natural logarithm shouldn’t frighten you: in exercises the text or the book supplies it, and you substitute. If $V_{2} > V_{1}$ the logarithm is positive and $L > 0$ .
Adiabatic: no heat exchanged, $Q = 0$ , so $Δ U = - L$ . If the gas expands it does work at the expense of its internal energy, and cools; if it is compressed it heats up. It is the reason a bicycle pump gets hot.

A quick example on the adiabatic: $1 mol$ of monatomic ideal gas expands without exchanging heat and the temperature drops by $20 K$ . Then $Δ U = 1 \cdot \frac{3}{2} \cdot 8.314 \cdot (- 20) \approx - 249 J$ , so $L = + 249 J$ : the gas has done $249 J$ of work, paying for it with its own internal energy.

Exercise 1: the isobaric

$2.0 mol$ of monatomic ideal gas are at $300 K$ and at the constant pressure of $1.0 \cdot 1 0^{5} Pa$ . They are heated to $450 K$ . Calculate $L$ , $Δ U$ and $Q$ .

Step 1: the temperature change. $Δ T = 450 - 300 = 150 K$ .

Step 2: the work. At constant pressure $L = n R Δ T$ :

L = 2.0 \cdot 8.314 \cdot 150 = 2494 J

Step 3: the change in internal energy.

Δ U = n \frac{3}{2} R Δ T = 2.0 \cdot 1.5 \cdot 8.314 \cdot 150 = 3741 J

Step 4: the heat, two ways. From the first law $Q = Δ U + L = 3741 + 2494 = 6235 J$ . As a check, by the isobaric formula $Q = n \frac{5}{2} R Δ T = 2.0 \cdot 2.5 \cdot 8.314 \cdot 150 = 6236 J$ (the one-joule difference is just rounding). The two ways agree: it is the check I recommend you always do.

Check on the volume. With $p V = n R T$ : $V_{1} = \frac{2.0 \cdot 8.314 \cdot 300}{1.0 \cdot 1 0 ^{5}} \approx 0.0499 m^{3}$ and $V_{2} \approx 0.0748 m^{3}$ , so $Δ V \approx 0.0249 m^{3}$ and $p Δ V = 2494 J$ . The gas has expanded (by about $25 L$ ), the work is positive, everything adds up. Note that only about $40%$ of the heat becomes work: the rest heats the gas.

To practise on timed questions there are the high school physics and maths tests: the correct answer is visible straight away, the guided step-by-step solution unlocks with your first lesson.

Exercise 2: the isothermal

$1.0 mol$ of ideal gas at $300 K$ expands isothermally from $10 L$ to $20 L$ . Calculate $Δ U$ , $L$ and $Q$ , and say whether the gas absorbs or gives off heat.

Step 1: internal energy. The temperature doesn’t change, so $Δ U = 0$ . No calculation needed.

Step 2: work. The volume ratio is $V_{2} / V_{1} = 2$ , and $ln 2 \approx 0.693$ :

L = n R T ln \frac{V _{2}}{V _{1}} = 1.0 \cdot 8.314 \cdot 300 \cdot 0.693 \approx 1729 J

The factor $n R T = 1.0 \cdot 8.314 \cdot 300 = 2494 J$ coincides by chance with the work in Exercise 1 ( $2 \cdot 150 = 1 \cdot 300$ ): it is not a rule, just a numerical equality of the data chosen.

Step 3: heat. $Δ U = Q - L = 0$ implies $Q = L \approx + 1729 J$ . The gas absorbs heat from the environment and turns all of it into work.

Check on the graph. On the $p$ - $V$ plane the isotherm is a falling hyperbola: $p_{1} = n R T / V_{1} = 2494/0.010 \approx 2.49 \cdot 1 0^{5} Pa$ and $p_{2} \approx 1.25 \cdot 1 0^{5} Pa$ . The area under the curve lies between the two rectangles: the smaller ( $p_{2} \cdot Δ V = 1.25 \cdot 1 0^{5} \cdot 0.010 = 1247 J$ ) and the larger ( $p_{1} \cdot Δ V = 2494 J$ ). The result $1729 J$ falls in between, as it should.

Exercise 3: a cycle

A monatomic ideal gas goes round the cycle $A \to B \to C \to D \to A$ on the $p$ - $V$ plane:

$A$ : $p = 1.0 \cdot 1 0^{5} Pa$ , $V = 0.010 m^{3}$ ;
$A \to B$ : isochoric up to $p = 2.0 \cdot 1 0^{5} Pa$ ;
$B \to C$ : isobaric up to $V = 0.030 m^{3}$ ;
$C \to D$ : isochoric down to $p = 1.0 \cdot 1 0^{5} Pa$ ;
$D \to A$ : isobaric down to $V = 0.010 m^{3}$ .

Calculate $L$ , $Δ U$ and $Q$ for each stage and the total work.

A trick that shortens the calculation: for a monatomic gas $n R Δ T = Δ (p V)$ , so $Δ U = \frac{3}{2} Δ (p V)$ and, in the isobaric stages, $Q = \frac{5}{2} Δ (p V)$ . You need neither the temperatures nor the number of moles.

Stage	$L$	$Δ U$	$Q$
$A \to B$ isochoric	$0$	$\frac{3}{2} (2000 - 1000) = + 1500 J$	$+ 1500 J$
$B \to C$ isobaric	$2.0 \cdot 1 0^{5} \cdot 0.020 = + 4000 J$	$\frac{3}{2} (6000 - 2000) = + 6000 J$	$+ 10, 000 J$
$C \to D$ isochoric	$0$	$\frac{3}{2} (3000 - 6000) = - 4500 J$	$- 4500 J$
$D \to A$ isobaric	$1.0 \cdot 1 0^{5} \cdot (- 0.020) = - 2000 J$	$\frac{3}{2} (1000 - 3000) = - 3000 J$	$- 5000 J$

Here $p V$ is: at $A$ $1000 J$ , at $B$ $2000 J$ , at $C$ $6000 J$ , at $D$ $3000 J$ . I obtained each $Q$ as $Δ U + L$ and checked it with the isobaric formula ( $\frac{5}{2} \cdot 4000 = 10, 000$ and $\frac{5}{2} \cdot (- 2000) = - 5000$ ).

Totals. $L_{t o t} = 4000 - 2000 = 2000 J$ ; $Δ U_{t o t} = 1500 + 6000 - 4500 - 3000 = 0$ ; $Q_{t o t} = 1500 + 10, 000 - 4500 - 5000 = 2000 J$ .

The three checks the cycle must pass: $Δ U_{t o t} = 0$ (internal energy is a state function, and at the end of the cycle you are back at the starting point); $Q_{t o t} = L_{t o t}$ (a consequence of the first law); and $L_{t o t}$ equal to the area of the rectangle, base $0.020 m^{3}$ times height $1.0 \cdot 1 0^{5} Pa$ , that is $2000 J$ , positive because the cycle runs clockwise ( $A \to B \to C \to D$ goes up, goes right, comes down, goes back left).

If the test asks for the efficiency, it is the useful work divided by the heat absorbed: $Q_{ab s} = 1500 + 10, 000 = 11, 500 J$ , so $η = 2000/11, 500 \approx 0.17$ , that is about $17%$ . More on efficiency in the second law, which usually follows straight after.

The mistakes that cost marks

The sign of the work. Expansion means $L > 0$ , compression $L < 0$ . With the convention of this post it is the last thing to check before handing in: did the gas expand or get compressed?
Celsius temperatures in the equation of state. $p V = n R T$ needs kelvin. With $27^{\circ} C$ you write $300 K$ , never $27$ .
Litres instead of cubic metres. If $V$ is in litres and $p$ in pascals, the work comes out a thousand times bigger or smaller. Convert $V$ to $m^{3}$ at the start.
$Δ U = Q$ everywhere. It holds only at constant volume. In an isobaric transformation part of the heat becomes work; in an isothermal one $Δ U = 0$ and $Q$ is all work.
Using $c_{P}$ for $Δ U$ . $Δ U$ is always calculated with $c_{V}$ , even in a constant-pressure transformation. $c_{P}$ is only for the heat of the isobaric.
Clockwise or anticlockwise cycle. If the cycle runs anticlockwise the total work is negative: the system has work done on it, and that is how a fridge works. Look at the direction of the arrows before writing the sign.
Forgetting that $Δ U$ of a cycle is zero. It is the free check that tells you whether you made a mistake in a stage: if the sum isn’t zero, look for the error before going on.

How to prepare

Redo the three exercises without looking, changing $n$ and the temperatures. Then try drawing each transformation on the $p$ - $V$ plane with arrows, and write next to each stage the three signs of $L$ , $Δ U$ , $Q$ : if the drawing and the signs are consistent, the calculations follow. If you still have basic thermal physics pending, after specific heat move on to changes of state and thermal expansion, and for those coming from a track with chemistry I recommend thermochemistry and enthalpy, where the first law reappears in a different notation. The overall picture of the subject is on the physics page.

If you want a check before the test

If the formulas make sense but the problems don’t, what is usually missing is a method: draw the $p$ - $V$ plane, write the signs of $L$ , $Δ U$ and $Q$ before the numbers, and use the first law as a final check. Write to me from the contact page and we’ll do a free introductory call: send me a thermodynamics exercise that doesn’t work out and I’ll tell you where it gets stuck.

First law of thermodynamics with worked exercises