Heat Capacity at Constant Volume

Equivalent forms

Q = n C_v \Delta T

C_v = \frac{f}{2} R

Equipartition makes a complicated quantity trivially countable — just count degrees of freedom and multiply by R/2.

Symbol	Name	SI unit	Dimension	Range
$Q$	Heat addedoutput Heat transferred to the system at constant V	J	M·L²·T⁻²	0 – 10000
$n$	Moles Number of moles of gas	mol	N	0.1 – 10
$C_v$	Molar heat capacity (const V) Energy per mole per K at constant volume	J/(mol·K)	M·L²·T⁻²·N⁻¹·Θ⁻¹	12 – 30
$ΔT$	Temperature change Change in temperature	K	Θ	1 – 500

Unit systems

Symbol	SI	CGS	Imperial
$Q$	J	erg	BTU
$n$	mol	mol	mol
$C_v$	J/(mol·K)	erg/(mol·K)	BTU/(mol·°R)
$ΔT$	K	K	°R

Where it holds

Equipartition

(C_v = fR/2)

valid when

k_B\cdot T

≫

\hbar \omega of

vibrational modes. Fails at low T where quantum effects freeze out modes.

Dimensional analysis

$[Q] = [n][C_v][\Delta T] = (N)(M\cdot L^{2}\cdot T^{-2}\cdot N^{-1}\cdot \Theta ^{-1})(\Theta ) = M\cdot L^{2}\cdot T^{-2} \checkmark$

Discovery

Joseph Black & James Joule · 1761

Black first distinguished heat from temperature; Joule's mechanical-equivalent experiments and Clausius's energy formulation in 1850 unified heat capacities into the thermodynamic definition.

Try this

Why does a steel pot heat faster than the water inside it?

1.5 mol of monatomic He gas is heated at constant volume. How much heat raises T by 40 K?

Research status: stable

Real-world applications

Bomb calorimetry (combustion at constant V)
Specific heat measurements in materials science
Atmospheric thermodynamics — heating of trapped air parcels
Cryogenic engineering

Common misconceptions

C_v is not the heat capacity of an isolated system — it's defined for a process at constant V
For solids and liquids, $C_v \approx C_p$ because volume changes negligibly, but the distinction matters for gases
Equipartition fails at low T — Einstein and Debye models explain $T^{3}$ behavior of solids

Experimental verification

Heat capacities measured via adiabatic calorimetry agree with equipartition for noble gases (3R/2) and diatomic gases at room T (5R/2) to within 1%.

Derivation

Start with first law:

dU = \delta Q - \delta W = \delta Q - PdV

.

At constant V,

dV = 0

,

so \delta Q = dU

.

Then

C_v = (\delta Q/dT)_V = (\partial U/\partial T)_V

.

For an ideal gas U depends only on T, so this is exact.

Limiting cases

Monatomic ideal gas⟶

C_v = 3R/2

Three translational degrees of freedom only.

Diatomic gas (room T)⟶

C_v = 5R/2

Three translation + two rotation modes; vibrational modes frozen out.

Solid (Dulong-Petit)⟶

C_v \approx 3R

Each atom oscillates in 3D — 6 quadratic terms in Hamiltonian

\times R/2

.

What if…

What if you double the moles?

You need twice the heat for the same $\Delta T$ — heat capacity is extensive.

What if vibrational modes activate (high T)?

C_v jumps to 7R/2 for diatomic gases — equipartition adds 2 more quadratic terms per vibrational mode.

What if

T \to 0

?

Quantum effects freeze degrees of freedom; $C_v \to 0$ as $T^{3} for$ solids (Debye law) and exponentially for gases.

1

Heat for 1.5 mol He, ΔT = 40 K

Given ·

n:: 1.5
C v:: 12.47
ΔT:: 40

Find · Q

Steps

He is monatomic $\to C_v = 3R/2 = 1.5 \times 8.314 = 12.47\,\mathrm{J}/(mol\cdot K)$
$Q = nC_v\Delta T = 1.5 \times 12.47 \times 40$
$Q \approx 748\,\mathrm{J}$

Result ·

Q = nC_v\Delta T = 1.5 \times 12.47 \times 40 = 748\,\mathrm{J}

2

ΔT of diatomic N₂ given Q

Given ·

n:: 2
Q:: 1000
C v:: 20.785

Find ·

\Delta T

Steps

$N_{2}$ diatomic $\to C_v = 5R/2 = 2.5 \times 8.314 = 20.785\,\mathrm{J}/(mol\cdot K)$
Rearrange: $\Delta T = Q/(nC_v)$
$\Delta T = 1000 / (2 \times 20.785) \approx 24.06\,\mathrm{K}$

Result ·

\Delta T = Q/(nC_v) = 1000/(2\times 20.785) = 24.06\,\mathrm{K}

Heat Capacity at Constant Pressure→First Law of Thermodynamics→Ideal Gas Law→