Class 12 > Unit # 16:First Law of Thermodynamics > Specific Heat & Relation Between Molar Specific Heats

Class 11 Physics • Thermodynamics

Specific Heat Capacity & Molar Heat Capacity

Definitions, determination by Method of Mixtures, Molar Specific Heat at constant pressure and volume, and step-by-step derivation of Mayer's Relation ($C_p - C_v = R$).

1. Specific Heat Capacity & Determination

Specific Heat Capacity: It is defined as the amount of heat required to raise the temperature of unit mass of the substance by unit degree rise of temperature.

If a body of mass $m$ absorbs heat $\Delta Q$ such that its temperature rises by $\Delta T$, then its total heat capacity $C'$ is:

$$C' = \frac{\Delta Q}{\Delta T}$$

Whereas the Specific Heat Capacity ($c$) per unit mass is given by:

$$c = \frac{1}{m} \left( \frac{\Delta Q}{\Delta T} \right) \implies \Delta Q = m \cdot c \cdot \Delta T$$

Determination of Specific Heat Capacity (Method of Mixtures)

The method of mixtures is the fundamental technique used to measure the specific heat capacity of solids and liquids. It operates on the Law of Heat Exchange:

$$\text{Heat Lost by Hot Bodies} = \text{Heat Gained by Cold Bodies}$$

A hot substance of known mass and temperature is mixed with a cooler liquid inside a insulated vessel called a calorimeter. Heat transfers until both reach a common final thermal equilibrium temperature, allowing the unknown specific heat to be calculated from the energy balance.

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2. Molar Specific Heat & Mayer's Relation ($C_p - C_v = R$)

Molar Specific Heat: Quantity of heat required to raise the temperature of one mole of a substance through $1\text{ K}$. Its SI unit is $\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.

If $n$ is the number of moles and $M$ is the molar mass, then mass $m = n \cdot M$. Substituting into the heat equation:

$$\Delta Q = (n \cdot M) \cdot c \cdot \Delta T = n \cdot (M \cdot c) \cdot \Delta T$$

$$\text{Since } C = M \cdot c \implies \Delta Q = n \cdot C \cdot \Delta T$$

Types of Molar Specific Heat in Gases

1. At Constant Volume ($C_v$)

Heat required to raise the temperature of one mole of gas through $1\text{ K}$ while keeping its volume fixed ($V = \text{Constant}$). No mechanical work is performed.

2. At Constant Pressure ($C_p$)

Heat required to raise the temperature of one mole of gas through $1\text{ K}$ while maintaining constant pressure ($P = \text{Constant}$). Gas expands and performs external work.

Derivation Proof: $C_p - C_v = R$

Step 1: Process at Constant Pressure
From the First Law of Thermodynamics ($\Delta Q = \Delta U + P\Delta V$), for $1\text{ mole}$ of gas at constant pressure:

$$\Delta Q_p = C_p \Delta T$$ $$C_p \Delta T = \Delta U + P\Delta V \quad \text{--- (i)}$$

Step 2: Process at Constant Volume
At constant volume, $\Delta V = 0 \implies \Delta W = 0$. Therefore, all supplied heat increases internal energy:

$$\Delta Q_v = C_v \Delta T = \Delta U \quad \text{--- (ii)}$$

Step 3: Combining Equations
Substitute equation (ii) into equation (i):

$$C_p \Delta T = C_v \Delta T + P\Delta V$$ $$(C_p - C_v)\Delta T = P\Delta V$$

Step 4: Applying Ideal Gas Equation
For $1\text{ mole}$ of an ideal gas ($PV = RT$), differentiating at constant pressure gives $P\Delta V = R\Delta T$. Substituting this back:

$$(C_p - C_v)\Delta T = R\Delta T$$

Dividing both sides by $\Delta T$ yields Mayer's Relation:

$$C_p - C_v = R$$

Where $R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ (General Gas Constant).

Key Conclusion: Since $R > 0$, it strictly holds that $C_p > C_v$. Heat energy supplied at constant pressure must both increase internal energy and do external work against expansion.