Heat Capacity at Constant Pressure

Equivalent forms

C_p = C_v + R

Q = n C_p \Delta T

C_p - C_v = R for any ideal gas — a single universal constant captures the cost of expansion work.

Symbol	Name	SI unit	Dimension	Range
$Q$	Heat addedoutput Heat transferred at constant P	J	M·L²·T⁻²	0 – 20000
$n$	Moles Number of moles of gas	mol	N	0.1 – 10
$C_p$	Molar heat capacity (const P) Energy per mole per K at constant pressure	J/(mol·K)	M·L²·T⁻²·N⁻¹·Θ⁻¹	20 – 40
$ΔT$	Temperature change Change in temperature	K	Θ	1 – 500

Unit systems

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

Where it holds

Valid for ideal gases with constant pressure process. For real gases, C_p - C_v deviates from R near the critical point.

Dimensional analysis

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

Discovery

Julius von Mayer · 1842

Mayer was the first to predict the difference C_p - C_v = R from energy conservation, four years before Joule's mechanical-equivalent measurement made it indisputable.

Try this

Why does an open balloon take more heat than a sealed one to warm up?

Heat 2 mol of an ideal diatomic gas from 300 K to 400 K at constant pressure. How much heat is needed?

Research status: stable

Real-world applications

HVAC and air-conditioning calculations
Jet engine combustion chamber design
Atmospheric thermodynamics (dry adiabatic lapse rate)
Speed of sound: $c = \sqrt{\gamma RT/M}$

Common misconceptions

C_p > C_v not because 'gas does work', but because at constant P the gas must expand, and dH already includes the PdV work
Mayer's relation only holds for ideal gases — for real gases it's modified by $(\partial U/\partial V)_T$
The ratio $\gamma = C_p/C_v$ governs adiabatic processes, not C_p alone

Experimental verification

Mayer's relation

C_p - C_v = R

confirmed for noble gases to within 0.5% at standard conditions. For air at

25^{\circ}C

:

C_p = 29.1

,

C_v = 20.8

→ difference

8.3 \approx R = 8.314

.

Derivation

Enthalpy

H = U + PV

.

At constant P,

\delta Q = dH

, so

C_p = (\partial H/\partial T)_P

.

For ideal gas,

H = U + nRT

, so

(\partial H/\partial T) = (\partial U/\partial T) + nR \to C_p = C_v + R

per mole (Mayer's relation).

Limiting cases

Monatomic ideal gas⟶

C_p = 5R/2

C_v = 3R/2

plus R for expansion work.

Diatomic gas (room T)⟶

C_p = 7R/2

C_v = 5R/2

plus R for expansion work.

Incompressible solid/liquid⟶

C_p \approx C_v

Volume barely changes so PdV work is negligible.

What if…

What if you heat the same gas at constant V instead?

You'd need only $Q = nC_v\Delta T$ — less heat because no expansion work is done.

What if the gas is polyatomic (e.g.,

CO_{2})

?

C_p rises $(\approx 37\,\mathrm{J}/(mol\cdot K)$ for $CO_{2}$ near room T) because more vibrational modes contribute.

What if pressure isn't constant?

C_p doesn't apply; you must integrate $dH = C_p dT + (\partial H/\partial P)_T dP$ .

1

Heat 2 mol N₂ from 300 K to 400 K at constant P

Given ·

n:: 2
C p:: 29.1
ΔT:: 100

Find · Q

Steps

C_p for diatomic $\approx 7R/2 = 29.1\,\mathrm{J}/(mol\cdot K)$
$Q = 2 \times 29.1 \times 100$
$Q \approx 5820\,\mathrm{J}$

Result ·

Q = nC_p\Delta T = 2 \times 29.1 \times 100 = 5820\,\mathrm{J}

2

Verify Mayer's relation for air

Given ·

C p:: 29.1
C v:: 20.8

Find · C_p - C_v

Steps

Subtract: $29.1 - 20.8 = 8.3$
Compare to $R = 8.314\,\mathrm{J}/(mol\cdot K)$
Agreement within 0.2%

Result ·

29.1 - 20.8 = 8.3\,\mathrm{J}/(mol\cdot K) \approx R = 8.314 \checkmark

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