8.9 Molar Specific Heats $C_p$ and $C_v$ | Heat And Thermodynamics | Physics 11

Heat Capacity and Specific Heat — Quick Review

Heat capacity ( $C$ ) is the heat required to raise the temperature of an object by 1 K: $C = \frac{Δ Q}{Δ T} (unit: J/K)$

Specific heat capacity ( $c$ ) accounts for mass: $Δ Q = c m Δ T ⟹ c = \frac{Δ Q}{m Δ T} (unit: J\cdotkg^{- 1} \cdotK^{- 1})$

Molar specific heat capacity ( $C_{m}$ ) uses moles instead of mass: $Δ Q = C_{m} n Δ T ⟹ C_{m} = \frac{Δ Q}{n Δ T} (unit: J\cdotmol^{- 1} \cdotK^{- 1})$

Molar Specific Heats of Gases: $C_{p}$ and $C_{v}$

For gases, the heat required for a given temperature change depends on whether the process occurs at constant volume or constant pressure.

Molar Specific Heat at Constant Volume ( $C_{v}$ )

$C_{v}$ is the heat required to raise the temperature of one mole of a gas by 1 K at constant volume.

First Law Application: At constant volume, $Δ V = 0$ , so $W = P Δ V = 0$ . The first law $Q = Δ U + W$ simplifies to: $Δ Q_{v} = Δ U ⟹ n C_{v} Δ T = Δ U$

All heat supplied goes entirely into increasing the internal energy of the gas.

Molar Specific Heat at Constant Pressure ( $C_{p}$ )

$C_{p}$ is the heat required to raise the temperature of one mole of a gas by 1 K at constant pressure.

First Law Application: At constant pressure, the gas expands and does work $W = P Δ V$ . The first law gives: $Δ Q_{p} = Δ U + W = Δ U + P Δ V ⟹ n C_{p} Δ T = Δ U + P Δ V$

Heat is used for two purposes: increasing internal energy AND doing expansion work.

Why is $C_{p} > C_{v}$ ?

Process	$Δ Q$	$Δ U$	$W$
Constant Volume	$n C_{v} Δ T$	$Δ U$	$0$
Constant Pressure	$n C_{p} Δ T$	$Δ U$ (same)	$P Δ V > 0$

At constant volume: all heat → internal energy only.
At constant pressure: heat → internal energy plus expansion work.
Therefore, more heat is needed at constant pressure for the same $Δ T$ , so $C_{p} > C_{v}$ .

Mayer's Relation: $C_{p} - C_{v} = R$

For an ideal gas, the difference between molar specific heats equals the Universal Gas Constant $R$ .

Derivation:

First law at constant pressure: $Δ Q_{p} = Δ U + W$
Substitute: $n C_{p} Δ T = n C_{v} Δ T + P Δ V$
From the ideal gas law at constant pressure: $P Δ V = n R Δ T$
Substitute: $n C_{p} Δ T = n C_{v} Δ T + n R Δ T$
Divide by $n Δ T$ : $C_{p} - C_{v} = R$

where $R \approx 8.314 J \cdot m o l^{- 1} \cdot K^{- 1}$ .

Physical meaning: The extra heat $R$ per mole per kelvin at constant pressure is exactly the work done by the gas expanding against constant pressure. This is a direct expression of the conservation of energy (First Law).

Key Questions

Q: What does $C_{p} - C_{v} = R$ physically mean?

The extra heat needed to raise 1 mole of an ideal gas by 1 K at constant pressure (compared to constant volume) equals the work done by the gas expanding: $W = P Δ V = n R Δ T$ . Per mole per kelvin, this is exactly $R$ .

Q: Does $C_{p} > C_{v}$ hold for solids and liquids?

Yes, but the difference is negligible. Solids and liquids expand very little when heated, so $P Δ V \approx 0$ and $C_{p} \approx C_{v}$ . We therefore use a single specific heat for solids and liquids.

Heat Capacity and Specific Heat — Quick Review

Heat capacity ( $C$ ) is the heat required to raise the temperature of an object by 1 K: $C = \frac{Δ Q}{Δ T} (unit: J/K)$

Specific heat capacity ( $c$ ) accounts for mass: $Δ Q = c m Δ T ⟹ c = \frac{Δ Q}{m Δ T} (unit: J\cdotkg^{- 1} \cdotK^{- 1})$

Molar specific heat capacity ( $C_{m}$ ) uses moles instead of mass: $Δ Q = C_{m} n Δ T ⟹ C_{m} = \frac{Δ Q}{n Δ T} (unit: J\cdotmol^{- 1} \cdotK^{- 1})$

Molar Specific Heats of Gases: $C_{p}$ and $C_{v}$

For gases, the heat required for a given temperature change depends on whether the process occurs at constant volume or constant pressure.

Molar Specific Heat at Constant Volume ( $C_{v}$ )

$C_{v}$ is the heat required to raise the temperature of one mole of a gas by 1 K at constant volume.

First Law Application: At constant volume, $Δ V = 0$ , so $W = P Δ V = 0$ . The first law $Q = Δ U + W$ simplifies to: $Δ Q_{v} = Δ U ⟹ n C_{v} Δ T = Δ U$

All heat supplied goes entirely into increasing the internal energy of the gas.

Molar Specific Heat at Constant Pressure ( $C_{p}$ )

$C_{p}$ is the heat required to raise the temperature of one mole of a gas by 1 K at constant pressure.

First Law Application: At constant pressure, the gas expands and does work $W = P Δ V$ . The first law gives: $Δ Q_{p} = Δ U + W = Δ U + P Δ V ⟹ n C_{p} Δ T = Δ U + P Δ V$

Heat is used for two purposes: increasing internal energy AND doing expansion work.

Why is $C_{p} > C_{v}$ ?

Process	$Δ Q$	$Δ U$	$W$
Constant Volume	$n C_{v} Δ T$	$Δ U$	$0$
Constant Pressure	$n C_{p} Δ T$	$Δ U$ (same)	$P Δ V > 0$

At constant volume: all heat → internal energy only.
At constant pressure: heat → internal energy plus expansion work.
Therefore, more heat is needed at constant pressure for the same $Δ T$ , so $C_{p} > C_{v}$ .

Mayer's Relation: $C_{p} - C_{v} = R$

For an ideal gas, the difference between molar specific heats equals the Universal Gas Constant $R$ .

Derivation:

First law at constant pressure: $Δ Q_{p} = Δ U + W$
Substitute: $n C_{p} Δ T = n C_{v} Δ T + P Δ V$
From the ideal gas law at constant pressure: $P Δ V = n R Δ T$
Substitute: $n C_{p} Δ T = n C_{v} Δ T + n R Δ T$
Divide by $n Δ T$ : $C_{p} - C_{v} = R$

where $R \approx 8.314 J \cdot m o l^{- 1} \cdot K^{- 1}$ .

Key Questions

Q: What does $C_{p} - C_{v} = R$ physically mean?

Q: Does $C_{p} > C_{v}$ hold for solids and liquids?