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Thermodynamics · Part 2 of 10

First Law of Thermodynamics

Heat a gas under a free piston and it gets hotter and pushes the piston up. The first law splits the heat into those two parts, $Δ Q = Δ U + Δ W$ , and this part shows how to compute each one: heat from a molar specific heat, work as an area under the P–V curve, and the internal-energy change from the temperature alone.

Builds on: Part 1 · Thermal Equilibrium and the Zeroth Law.

Video coming soon

First law

$Δ Q = Δ U + Δ W$

Heat supplied to a gas raises its internal energy and lets it do work on the surroundings. Signs: $Q > 0$ when heat goes into the gas, $W > 0$ when the gas does work (expands), $Δ U > 0$ when it gets hotter. For a tiny step, $d Q = d U + d W$ .

A gas in a cylinder with a piston, heated by a flame. — Heat in: hotter gas, piston pushed up.

P–V diagram with two paths and the areas under them. — Paths a and b join the same two states but do different work.

Work done by a gas

$W = \int P d V$

The gas pushes the piston (area S) with force $P S$ ; moving it by $d x$ does work $d W = P S d x = P d V$ . Over a process, $W = \int_{V_{1}}^{V_{2}} P d V$ : the area under the P–V curve. Different paths between the same states enclose different areas, so W depends on the path.

Constant pressure

$W = P Δ V = n R Δ T$

At constant pressure the area under the path is a rectangle: $W = P (V_{2} - V_{1}) = n R (T_{2} - T_{1})$ . For 1 mol heated by 100 K at 1 atm, $W \approx 831$ J.

Rectangle under a horizontal isobar. — 1 mol at 1 atm, 22.4 L → 30.6 L.

Internal energy

$Δ U = \frac{f}{2} n R Δ T$

For an ideal gas $U = \frac{f}{2} n R T$ depends only on temperature, so $Δ U = \frac{f}{2} n R (T_{2} - T_{1})$ in every process. Molar specific heat C: $d Q = n C d T$ , and a gas's C depends on the process.

Summary

Key formulas

d Q = d U + d W

Statement and equation of the first law

Δ Q = Δ U + Δ W

Statement and equation of the first law

d Q = n C d T

Specific Heat Capacities of Gases

Q = n C (T_{2} - T_{1})

Specific Heat Capacities of Gases

d W = P d V

Heating of a Gaseous System and Work done by a Gas

W = \int_{V_{1}}^{V_{2}} P d V

Heating of a Gaseous System and Work done by a Gas

d U = \frac{f}{2} n R d T

Change in Internal Energy of a Gas on Heating

Δ U = \frac{f}{2} n R (T_{2} - T_{1})

Change in Internal Energy of a Gas on Heating

Worked examples

One for every idea

Statement and equation of the first law

1. In a process the internal energy of a system decreases by 400 J while the system does 250 J of work on its surroundings. What net heat is taken in by the system?

Signs: $Δ U = - 400$ J (it falls), $W = + 250$ J (done by the system).
$Q = Δ U + W = - 400 + 250$ .

$Q = - 150$ J: the system actually rejects 150 J of heat. (Illustrative Example 3.2)

Specific Heat Capacities of Gases

2. Find the heat required to raise the temperature of 2 mol of a diatomic gas by 50 K at constant volume, given $C_{v} = \frac{5}{2} R$ .

$C_{v} = \frac{5}{2} \times 8.314 = 20.785$ J/mol·K.
$Q = n C_{v} Δ T = 2 \times 20.785 \times 50$ .

$Q \approx 2079$ J (about 2.08 kJ).

Heating of a Gaseous System and Work done by a Gas

3. One mole of an ideal gas is heated from 0 °C to 100 °C at a constant pressure of 1 atm. Find the work done by the gas.

At constant pressure $W = P (V_{2} - V_{1}) = n R Δ T$ .
A step of 100 °C is a step of 100 K: $W = 1 \times 8.314 \times 100$ .
Check with volumes: 22.4 L → 30.6 L, $W = 1.013 \times 1 0^{5} \times 0.0082 \approx 831$ J.

$W \approx 831$ J. (Illustrative Example 3.1; the book's 830.7 J comes from rounding the volumes.)

Change in Internal Energy of a Gas on Heating

4. Hydrogen ( $f = 5$ ) at 273 K and $1 0^{5}$ Pa fills $5 \times 1 0^{- 3}$ m³. It is cooled by 55 K at constant volume. Find the change in internal energy and the heat lost.

$n = \frac{P V}{R T} = \frac{1 0 ^{5} \times 5 \times 1 0 ^{- 3}}{8.314 \times 273} \approx 0.22$ mol.
$Δ U = \frac{5}{2} n R Δ T = 2.5 \times 0.22 \times 8.314 \times (- 55) \approx - 252$ J.
Constant volume: $W = 0$ , so $Q = Δ U$ .

$Δ U \approx - 252$ J and the gas loses about 252 J of heat. (Illustrative Example 3.5 gives 251.5 J with n rounded to 0.22; note the book's "STP" here is $1 0^{5}$ Pa.)

JEE-style question

Your turn

A gas absorbs 200 J of heat and does 80 J of work on its surroundings. Its internal energy:

(a)increases by 280 J

(b)increases by 120 J

(c)decreases by 120 J

(d)increases by 200 J

Show the answer and the traps

ΔU = Q − W = 200 − 80 = 120 J. It increases by 120 J: option b.

280 J adds the work instead of subtracting it. A decrease flips the sign.

And 200 J forgets that part of the heat left as work.

Watch out

Common mistakes

Assuming a gas has one fixed specific heatC depends on the process:

C_{v}

at constant volume,

C_{p}

at constant pressure, others in between.

Thinking ΔU depends on the pathU depends only on T:

Δ U = \frac{f}{2} n R Δ T

for any path between the same temperatures.

Mixing up signsHeat given out is

Q < 0

; work done on the gas is

W < 0

Practice

Try these

1. A gas absorbs 500 J of heat and its internal energy rises by 300 J. Find the work done by the gas.

$W = Q - Δ U = 200$ J.

2. Find the heat needed to raise 3 mol of a monatomic gas (

C_{v} = \frac{3}{2} R

) by 20 K at constant volume.

$Q = 3 \times 12.47 \times 20 \approx 748$ J.

3. 2 mol of an ideal gas is heated at constant pressure from 300 K to 350 K. Find the work done by the gas.

$W = n R Δ T = 2 \times 8.314 \times 50 \approx 831$ J.

4. The temperature of 4 mol of a diatomic gas (

f = 5

) falls from 400 K to 350 K. Find ΔU.

$Δ U = 2.5 \times 4 \times 8.314 \times (- 50) \approx - 4157$ J.

First Law of Thermodynamics

ΔQ=ΔU+ΔW

W=∫PdV

W=PΔV=nRΔT

ΔU=2f​nRΔT

Key formulas

One for every idea

Your turn

Common mistakes

Try these

$Δ Q = Δ U + Δ W$

$W = \int P d V$

$W = P Δ V = n R Δ T$

$Δ U = \frac{f}{2} n R Δ T$