Laws of Thermodynamics

Lessons

Senior 4 · Physics

Laws of Thermodynamics

5 lessons

Lesson 2 of 5

The First Law of Thermodynamics: Isochoric, Isobaric and Isothermal Changes

25 min

Falling weights and warm water

In the 1840s James Joule let weights fall slowly, turning a paddle wheel inside an insulated can of water. The stirring warmed the water, very slightly but measurably, and the rise in temperature was always proportional to the work done by the falling weights, whoever did the experiment and however it was arranged. He had shown that heating and doing work are two ways of transferring the same thing, energy, and he measured the exchange rate: about 4.2 J of work raises the temperature of a gram of water as much as one calorie of heat. This lesson turns his result into the first law of thermodynamics and applies it to the simplest changes a gas can make.

The first law

Energy is conserved. If heat $Q$ is supplied to a system and the system does work $W$ on its surroundings, its internal energy changes by

\Delta U = Q - W

The signs matter, so fix them once:

Quantity	Positive when	Negative when
$Q$	heat flows into the system	heat flows out of the system
$W$	the system does work (expands)	work is done on the system (it is compressed)
$\Delta U$	internal energy rises (an ideal gas warms)	internal energy falls (it cools)

Table 2.1: The sign convention used in this course. Some books write $\Delta U = Q + W$ , where $W$ means the work done on the system; the physics is the same.

The first law says that you cannot get work out of nothing: an engine that delivered work without taking in heat or using up internal energy would break it.

Definition

The first law of thermodynamics: the increase in internal energy of a system equals the heat supplied to it minus the work done by it, $\Delta U = Q - W$ . It is the law of conservation of energy applied to heat and work. Example: a gas absorbs 500 J of heat and does 200 J of work pushing a piston, so $\Delta U = 500 - 200 = +300\,\mathrm{J}$ ; its temperature rises.

Worked example 1: Joule's warm water

In an experiment like Joule's, a 2.0 kg mass falls 1.5 m twenty times, turning paddles in 0.50 kg of water in an insulated can. Assuming all the work goes into the water ( $c = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$ ), find the rise in temperature.

W = 20\,mgh = 20 \times 2.0 \times 9.8 \times 1.5 = 588\,\mathrm{J}, \qquad \Delta T = \frac{W}{mc} = \frac{588}{0.50 \times 4200} = 0.28\,\mathrm{K}

With $Q = 0$ (insulated can) and work done on the water ( $W = -588\,\mathrm{J}$ ), the first law gives $\Delta U = +588\,\mathrm{J}$ . The rise is less than a third of a degree, which is why Joule needed very precise thermometers.

Three simple changes of an ideal gas

For $n$ moles of ideal gas, define the molar heat capacities: $C_{V}$ is the heat needed to warm 1 mol by 1 K at constant volume, and $C_{p}$ the heat needed at constant pressure. Because $U$ depends only on temperature, $\Delta U = nC_V\Delta T$ for any change of an ideal gas.

Isochoric change (constant volume). The gas cannot push anything, so $W = 0$ and $\Delta U = Q = nC_V\Delta T$ : all the heat raises the internal energy. For a monatomic gas, $U = \tfrac{3}{2}nRT$ gives $C_V = \tfrac{3}{2}R$ .

Isobaric change (constant pressure). The gas expands as it warms, doing work $W = p\Delta V = nR\Delta T$ (from $p V = n R T$ ). The heat needed is $Q = nC_p\Delta T$ , so the first law gives

nC_p\Delta T = nC_V\Delta T + nR\Delta T \qquad\Rightarrow\qquad C_p - C_V = R

More heat is needed at constant pressure, because some of it leaves again as work. For a monatomic gas $C_p = \tfrac{5}{2}R$ ; for air, whose molecules are mostly diatomic, $C_V \approx \tfrac{5}{2}R$ and $C_p \approx \tfrac{7}{2}R$ . The ratio $\gamma = \frac{C_p}{C_V}$ is $\frac{5}{3}$ for a monatomic gas and about 1.40 for air; you need it in Lesson 3.

Isothermal change (constant temperature). If a gas expands slowly in good contact with a large body at a fixed temperature, $T$ stays constant, so $\Delta U = 0$ and $Q = W$ : every joule of heat that flows in leaves as work. The gas follows $pV = \text{constant}$ , a curve called an isotherm, and the work (the area under it) is

W = nRT\ln\frac{V_2}{V_1}

Figure 2.1: Isobaric, isochoric and isothermal changes from the same starting state S. Isotherms further from the origin belong to higher temperatures.

Change	Constant	$W$	$\Delta U$	$Q$
isochoric	$V$	0	$nC_V\Delta T$	$nC_V\Delta T$
isobaric	$p$	$p\Delta V = nR\Delta T$	$nC_V\Delta T$	$nC_p\Delta T$
isothermal	$T$	$nRT\ln\frac{V_2}{V_1}$	0	$W$

Table 2.2: The first law applied to three changes of $n$ moles of ideal gas.

Definition

An isochoric change happens at constant volume, an isobaric change at constant pressure, and an isothermal change at constant temperature. The molar heat capacities $C_{V}$ and $C_{p}$ are the heat needed per mole per kelvin at constant volume and at constant pressure; for an ideal gas $C_{p} - C_{V} = R$ and $\gamma = \frac{C_p}{C_V}$ . Example: for helium, $C_V = \tfrac{3}{2}R = 12.5\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $C_p = \tfrac{5}{2}R = 20.8\,\mathrm{J\,mol^{-1}\,K^{-1}}$ .

Worked example 2: heating at constant pressure, and at constant volume

2.0 mol of a monatomic ideal gas is heated from 300 K to 350 K. Find $\Delta U$ , $W$ and $Q$ (a) at constant pressure, (b) at constant volume.

(a) $\Delta U = nC_V\Delta T = 2.0 \times \tfrac{3}{2} \times 8.31 \times 50 = 1.25 \times 10^{3}\,\mathrm{J}$ ; $W = nR\Delta T = 2.0 \times 8.31 \times 50 = 831\,\mathrm{J}$ ; $Q = \Delta U + W = 2.08 \times 10^{3}\,\mathrm{J}$ (check: $nC_p\Delta T = 2.0 \times \tfrac{5}{2} \times 8.31 \times 50 = 2.08 \times 10^{3}\,\mathrm{J}$ ).

(b) $W = 0$ , so $Q = \Delta U = 1.25 \times 10^{3}\,\mathrm{J}$ . The same temperature rise costs 40% less heat when the gas is not allowed to expand.

Worked example 3: an isothermal expansion

0.50 mol of ideal gas at 300 K expands isothermally from 10 L to 30 L. Find the work done, the heat absorbed and the change in internal energy.

W = nRT\ln\frac{V_2}{V_1} = 0.50 \times 8.31 \times 300 \times \ln 3 = 1.37 \times 10^{3}\,\mathrm{J}

$\Delta U = 0$ because $T$ is constant, so $Q = W = 1.37 \times 10^{3}\,\mathrm{J}$ : the gas takes in 1.37 kJ of heat and turns all of it into work. (Lesson 4 shows why no engine can keep doing this for ever.)

Warning

Common mistakes. (1) Mixing sign conventions: decide whether $W$ is work done by or on the gas, and keep to it. (2) Thinking that "isothermal" means "no heat flows": heat does flow, exactly enough to replace the work done. (3) Using $\Delta U = nC_p\Delta T$ : for an ideal gas $\Delta U = nC_V\Delta T$ in every change; $C_{p}$ is only for the heat at constant pressure. (4) Using degrees Celsius in $p V = n R T$ or $nRT\ln\frac{V_2}{V_1}$ : temperatures must be in kelvin (a temperature difference is the same in both).

Tip

One law, three short cuts. Write $\Delta U = Q - W$ every time, then cross out what the process makes zero: isochoric $W = 0$ ; isothermal $\Delta U = 0$ ; (next lesson) adiabatic $Q = 0$ . For an ideal gas, $\Delta U = nC_V\Delta T$ always.

The First Law of Thermodynamics: Internal Energy, Heat, and Work (Professor Dave Explains). Watch how heat and work both change the internal energy. Check which sign convention for work the video uses; in this course W is the work done by the gas, so ΔU = Q − W.

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Practice

A gas absorbs 800 J of heat and does 300 J of work. Find the change in its internal energy.
250 J of work is done on a gas while it gives out 100 J of heat. Find the change in its internal energy, and say whether its temperature rises or falls.
3.0 mol of a monatomic ideal gas is heated at constant volume through 20 K. Find $Q$ , $W$ and $\Delta U$ .
1.0 mol of air ( $C_V = \tfrac{5}{2}R$ ) is heated at constant pressure through 20 K. Find $Q$ , $\Delta U$ and $W$ .
2.0 mol of ideal gas at 400 K is compressed isothermally to half its volume. Find the work done by the gas and the heat that flows, and state the direction of the heat flow.
Starting from the first law and $p V = n R T$ , show that $C_{p} - C_{V} = R$ for an ideal gas.
Explain why more heat is needed to warm a gas by 1 K at constant pressure than at constant volume.
A student writes: "In an isothermal expansion no heat flows, because the temperature does not change." Evaluate this statement.

Show answersHide answers

$\Delta U = 800 - 300 = +500\,\mathrm{J}$ .
$Q = -100\,\mathrm{J}$ , $W = -250\,\mathrm{J}$ : $\Delta U = -100 - (-250) = +150\,\mathrm{J}$ ; the temperature rises.
$W = 0$ ; $Q = \Delta U = 3.0 \times \tfrac{3}{2} \times 8.31 \times 20 = 748\,\mathrm{J}$ .
$Q = nC_p\Delta T = 1.0 \times \tfrac{7}{2} \times 8.31 \times 20 = 582\,\mathrm{J}$ ; $\Delta U = \tfrac{5}{2} \times 8.31 \times 20 = 416\,\mathrm{J}$ ; $W = nR\Delta T = 166\,\mathrm{J}$ (check: $416 + 166 = 582$ ).
$W = nRT\ln\frac{V_2}{V_1} = 2.0 \times 8.31 \times 400 \times \ln 0.5 = -4.6 \times 10^{3}\,\mathrm{J}$ (work is done on the gas); $\Delta U = 0$ , so $Q = W = -4.6\,\mathrm{kJ}$ : 4.6 kJ of heat flows out of the gas.
At constant pressure $Q = \Delta U + W$ : $nC_p\Delta T = nC_V\Delta T + p\Delta V$ , and $p\Delta V = nR\Delta T$ . Dividing by $n\Delta T$ : $C_{p} = C_{V} + R$ .
At constant volume all the heat raises the internal energy. At constant pressure the gas expands and does work on its surroundings, so extra heat is needed to supply that work as well as the same rise in internal energy.
Wrong. Constant temperature means $\Delta U = 0$ , not $Q = 0$ . By the first law $Q = W$ : an expanding gas does work, so an equal amount of heat must flow in to keep its temperature from falling.

Quick check

Key points

First law: $\Delta U = Q - W$ ( $Q$ into the system, $W$ done by it): energy is conserved.
For an ideal gas $\Delta U = nC_V\Delta T$ in every change; $C_{p} - C_{V} = R$ ; monatomic $C_V = \tfrac{3}{2}R$ , $C_p = \tfrac{5}{2}R$ , $\gamma = \frac{5}{3}$ ; air $\gamma \approx 1.40$ .
Isochoric: $W = 0$ , $Q = \Delta U$ . Isobaric: $W = p\Delta V = nR\Delta T$ , $Q = nC_p\Delta T$ .
Isothermal: $\Delta U = 0$ , $Q = W = nRT\ln\frac{V_2}{V_1}$ ; the gas follows $p V =$ constant.

Sources and credits

Image: Joule's Apparatus (Harper's Scan), engraving from Harper's New Monthly Magazine, vol. 39 (August 1869), Wikimedia Commons, public domain.
Video: The First Law of Thermodynamics: Internal Energy, Heat, and Work, Professor Dave Explains, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagram, tables, worked examples and practice exercises.

1Falling weights and warm water

2The first law

3Three simple changes of an ideal gas

Falling weights and warm water

The first law

Three simple changes of an ideal gas