PhysicsThermodynamicsJEE · NEET · NSEP · INPhO · IPhO

Thermodynamics · Part 5 of 10

Isochoric, Isobaric and Isothermal Processes

Hold one quantity fixed and the first law becomes simple. This part works through the isochoric (V fixed), isobaric (P fixed) and isothermal (T fixed) processes: each one's P–V curve and its Q, W and ΔU, always with $Δ U = n C_{v} Δ T$ .

Builds on: Part 3 · Indicator Diagrams, State Variables and Cycles, Part 4 · Molar Specific Heats: Cv, Cp and γ.

Video coming soon

Isochoric

V constant: $W = 0$

A vertical line on the P–V diagram, with no area under it, so $W = 0$ and $Q = Δ U = n C_{v} Δ T = \frac{n R ( T _{2} - T _{1} )}{γ - 1}$ . $P / T$ stays constant.

Vertical line on a P–V diagram between two isotherms. — From isotherm T₁ up to isotherm T₂.

Horizontal line with the rectangle beneath shaded. — Work = rectangle under the isobar.

Isobaric

P constant: $W = n R Δ T$

A horizontal line; the work is the rectangle under it, $W = P (V_{2} - V_{1}) = n R (T_{2} - T_{1})$ . Heat $Q = n C_{p} Δ T$ , more than the isochoric heat by exactly the work.

Isothermal

T constant: $Q = W$

$P V$ = constant: a rectangular hyperbola, higher for higher T; isotherms never cross. $Δ U = 0$ , so $Q = W = n R T ln \frac{V _{2}}{V _{1}} = n R T ln \frac{P _{1}}{P _{2}}$ .

Family of isotherm hyperbolas with the area under one segment. — Three isotherms; expansion along the middle one.

Comparison table of the three processes. — Q, W and ΔU for n moles.

Side by side

One table for all three

Isochoric: W = 0. Isobaric: W = nRΔT, Q = nC_pΔT. Isothermal: ΔU = 0, Q = W. In every process $Δ U = n C_{v} Δ T$ , then $Q = Δ U + W$ .

Summary

Key formulas

isochoric: W = 0

Isochoric or Isometric Process

Q = Δ U = n C_{v} Δ T = \frac{n R Δ T}{γ - 1}

Isochoric or Isometric Process

isobaric: W = n R (T_{2} - T_{1})

Isobaric Process

Q = n C_{p} Δ T

Isobaric Process

Δ U = n C_{v} Δ T (any process)

Isobaric Process

isothermal: P V = const

Isothermal Process

W = n R T ln \frac{V _{2}}{V _{1}} = n R T ln \frac{P _{1}}{P _{2}}

Isothermal Process

Q = W

Isothermal Process

Worked examples

One for every idea

Isochoric or Isometric Process

1. 2 mol of a monatomic gas is heated at constant volume from 300 K to 400 K. Find the heat supplied.

$Q = n C_{v} Δ T$ with $C_{v} = \frac{3}{2} R$ .
$Q = 2 \times \frac{3}{2} \times 8.314 \times 100$ .

$Q \approx 2494$ J, all into internal energy.

Isobaric Process

2. 1 mol of a diatomic gas expands at constant pressure from 300 K to 400 K. Find the work done by the gas.

$W = n R (T_{2} - T_{1})$ : no pressure or volume needed.
$W = 1 \times 8.314 \times 100$ .
(Q would be $\frac{7}{2} n R Δ T \approx 2910$ J.)

$W \approx 831$ J.

Isothermal Process

3. One mole of an ideal gas at 27 °C (300 K) expands isothermally until its volume doubles. Find the work done.

$W = n R T ln \frac{V _{2}}{V _{1}} = 8.314 \times 300 \times ln 2$ .
$Δ U = 0$ , so $Q = W$ .
Use ln: $lo g_{10} 2$ would give 751 J.

$W = Q \approx 1729$ J.

JEE-style question

Your turn

2 mol of an ideal gas expand isothermally at 300 K from 10 L to 20 L. The heat absorbed by the gas is about:

(a)0

(b)3458 J

(c)1502 J

(d)4988 J

Show the answer and the traps

Q = W = nRT ln 2 = 2 × 8.314 × 300 × 0.693 ≈ 3458 J: option b.

Zero is ΔU, not Q. 1502 J uses log base 10.

And 4988 J is just nRT: it forgets the ln 2.

Watch out

Common mistakes

Using log₁₀ in W = nRT ln(V₂/V₁)It is the natural log: ln 2 = 0.693, not 0.301.

Writing ΔU = nC_pΔT in an isobaric processΔU is always

n C_{v} Δ T

;

n C_{p} Δ T

is the heat Q.

Thinking an isothermal process has no heatΔU = 0, but Q = W ≠ 0: all the heat becomes work.

Practice

Try these

1. At constant volume the pressure of a gas at 300 K is 2 atm. Find the pressure at 450 K.

$P / T$ constant: 3 atm.

2. 3 mol of a diatomic gas is heated at constant pressure from 280 K to 320 K. Find W, ΔU and Q.

$W = 3 \times 8.314 \times 40 \approx 998$ J, $Δ U \approx 2494$ J, $Q \approx 3492$ J.

3. 1 mol of ideal gas at 400 K is compressed isothermally from 10 L to 5 L. Find the work done by the gas and the heat exchanged.

$W = 8.314 \times 400 \times ln 0.5 \approx - 2305$ J; $Q = W$ : about 2305 J of heat is released.

Isochoric, Isobaric and Isothermal Processes

V constant: W=0

P constant: W=nRΔT

T constant: Q=W

One table for all three

Key formulas

One for every idea

Your turn

Common mistakes

Try these

V constant: $W = 0$

P constant: $W = n R Δ T$

T constant: $Q = W$