__What are Specific Heats of gases?__

**Specific heat** of a gas is defined as the quantity of heat required to raise the temperature of unit mass of gas by 1 \degree C .

When a gas is heated, its pressure and volume both changes with increase in temperature. Hence, for a gaseous substance, the *specific heat* is not a constant. It ranges from a negative value to a positive value, zero and even infinity. This can be explained as below –

### (1) Zero Specific Heat of gas

Consider ( m ) * *mass of a gas enclosed in a piston cylinder arrangement. Suppose, the gas is suddenly compressed by moving the piston. Let, the temperature of the gas rise by the heat of compression by ( \Delta T ) .

Since, no *heat* is added to the system, so ( \Delta Q = 0 ) . Therefore, * specific heat of gas *will be –

c = \left ( \frac { \Delta Q }{ m \Delta T } \right ) = \left ( \frac { 0 }{ m \Delta T } \right ) = 0 .

*Therefore, specific heat in this situation is zero.*

### (2) Infinite Specific Heat of gas

Suppose the gas in piston cylinder arrangement is heated and allowed to expand. The *temperature* of the gas will rise due to addition of heat by ( \Delta T ) .

Again, the gas is expanded and due to expansion of gas its temperature will fall. Let, the fall in temperature due to expansion is also ( \Delta T ) .

Then, the net change in temperature is ( \Delta T = 0 ) . Therefore, * specific heat of gas *will be –

c = \left ( \frac { \Delta Q }{ m \Delta T } \right ) = \left ( \frac { \Delta Q }{ m \times 0 } \right ) = \infty

*Therefore, specific heat in this situation is infinite.*

### (3) Positive Specific Heat of gas

Suppose, the gas is heated and allowed to expand. The rate of expansion is such that, fall in temperature of the gas due to expansion is less than rise in temperature due to addition of heat.

Then, the net change in temperature is a positive value i.e. ( \Delta T > 1 ) . Therefore, * specific heat of gas *will be –

c = \left ( \frac { \Delta Q }{ m \Delta T } \right ) > 1 .

*Therefore, specific heat in this situation is a positive value.*

### (4) Negative Specific Heat of gas

Suppose, the gas is heated and allowed to expand. The rate of expansion is such that, fall in temperature of the gas due to expansion is more than rise in temperature due to addition of heat.

Then, the net change in temperature is a negative value ( \Delta T < 1 ) . Therefore, * specific heat of gas *will be –

c = \left ( \frac { \Delta Q }{ m \Delta T } \right ) < 1 .

*Therefore, specific heat in this situation is a negative value.*

__Principal Specific Heat__

Out of many specific heats of a gas, two are of most important which are called principal specific heats. These are –

- Specific heat at constant volume. It is denoted by ( c_v )
- Specific heat at constant pressure. It is denoted by ( c_p )

__Specific Heat at Constant Volume__

**Specific heat at constant volume** of a gas is defined as the quantity of heat required to raise the temperature of unit mass of a gas by 1 \degree C at constant volume.

It is denoted by ( c_v ) . If, ( \Delta Q ) is the quantity of heat added to a system at constant volume and ( \Delta T ) is the rise in temperature. Then, * specific heat* –

c_v = \left ( \frac {\Delta Q}{m \Delta T} \right )

** Molar specific heat at constant volume – **It is the

*specific heat*of a gas if one mole of the gas is taken. It is denoted by ( C_v )

Therefore, \quad C_v = \left ( \frac {\Delta Q}{n \Delta T} \right )

Here, \quad n = \left [ \frac {\text {Mass of gas}}{\text {Molecular mass}} \right ]

__Specific Heat at Constant Pressure__

** Specific heat at constant pressure** of a gas is defined as the quantity of heat required to raise the temperature of unit mass of a gas by 1 \degree C at constant pressure. It is denoted by ( c_p )

If, ( \Delta Q ) is the quantity of heat added to a system at constant pressure and ( \Delta T ) is the rise in temperature, then –

c_p = \left ( \frac {\Delta Q}{m \Delta T} \right )

** Molar specific heat at constant pressure – **It is the specific heat of a gas if one mole of the gas is taken. It is denoted by ( C_p )

Therefore, \quad C_p = \left ( \frac {\Delta Q}{n \Delta T} \right )

## Relation between two Specific Heats of gases

__Mayer’s Formula__

*The relation between molar specific heats is given by Mayer’s formula.*

Consider that, one mole of an ideal gas is heated to raise its temperature by ( \Delta T ) . By *first law of thermodynamics*, total heat added, ( \Delta Q ) is used in increasing the *internal energy* ( \Delta U ) and doing work of expansion.

Therefore, \quad \Delta Q = ( \Delta U + P \Delta V )

If heat is added at constant volume, then –

\Delta V = 0

Or, \quad \Delta Q = \Delta U ( At constant volume. )

Therefore, specific heat at constant volume of a gas –

C_v = \left ( \frac {\text {Heat added at constant volume}}{\text {Rise in temperature}} \right ) = \left ( \frac {\Delta Q}{\Delta T} \right )

Using relation from equation (1), we get –

C_v = \left ( \frac {\Delta U}{\Delta T} \right ) …….. (1)

If heat is added at constant pressure, then –

\Delta V \ne 0

Therefore, specific heat at constant pressure of a gas –

C_p = \left ( \frac {\Delta Q}{\Delta T} \right )

Or, \quad C_p = \left ( \frac {\Delta U}{\Delta T} \right ) + P \left ( \frac {\Delta V}{\Delta T} \right ) ………. (2)

Subtracting equations (1) from equation (2), we get –

C_p - C_v = P \left ( \frac {\Delta V}{\Delta T} \right ) ………. (3)

From *Ideal gas equation* for one mole of a gas, we know –

PV = RT

Differentiating both sides with respect to ( T ) taking pressure as constant, we have –

P \Delta V = R \Delta T

Therefore, at constant pressure –

P \left ( \frac {\Delta V}{\Delta T} \right ) = R

Putting this value in equation (3), we get –

( C_p - C_v ) = R ……… (4)

*This relation is called Mayer’s formula.*

__Adiabatic relation between Pressure & Volume__

An adiabatic relationship can be established between pressure and volume as follows –

From *first law of thermodynamics*, we get –

\Delta Q = ( \Delta U + P \Delta V )

For an *adiabatic process,* ( \Delta Q = 0 )

Therefore, \quad \Delta U + P \Delta V = 0

But, \quad \Delta U = C_v \Delta T

Therefore, \quad C_v \Delta T + P \Delta V = 0 ………. (1)

Also, from ideal gas equation, we get –

PV = RT

By partial differentiation of both sides, we get –

P \Delta V + V \Delta P = R \Delta T

Or, \quad \Delta T = \left [ \frac {P \Delta V + V \Delta P}{R} \right ] ………. (2)

From relation (1) and (2), we get –

C_v \left [ \frac { ( P \Delta V + V \Delta P ) }{R} \right ] + P \Delta V = 0

Or, \quad C_v P \Delta V + C_v V \Delta P + R P \Delta V = 0

So, \quad C_v V \Delta P + \left ( C_v + R \right ) P \Delta V = 0

But, \quad ( C_p - C_v ) = R

Therefore, \quad C_v V \Delta P + C_p P \Delta V = 0

Dividing all terms by ( C_v \ P V ) , we get –

\left ( \frac {\Delta P}{P} \right ) + \left ( \frac {C_p}{C_v} \right ) \left ( \frac {\Delta V}{V} \right ) = 0

Or, \quad \left ( \frac {\Delta P}{P} \right ) + \gamma \left ( \frac {\Delta V}{V} \right ) = 0

Because, \left [ \gamma = \left ( \frac {C_p}{C_v} \right ) \right ]

Integrating both sides, we get –

\int \left ( \frac {\Delta P}{P} \right ) + \gamma \int \left ( \frac {\Delta V}{V} \right ) = C

Where, ( C ) is an integration constant.

Therefore, ( \log_e P + \gamma \log_e V ) = C

Or, \quad \log_e PV ^ \gamma = C

So, \quad PV ^ \gamma = e^C

Hence, \quad PV ^ \gamma = \text {Constant}

This is the adiabatic relation between pressure ( P ) and volume ( V ) of an ideal gas.

See numerical problems based on this article.