Specific Heat Capacity

Specific Heat Capacities of Gases

S = $\frac{1}{m}\left[\frac{\Delta Q}{\Delta T}\right]$

where $\Delta$Q = amount of heat required for '$\Delta$T' temperature change.

m = mass of the gas.

In case of gases, the concept of a molar heat capacity is useful. Molar heat capacity is the amount of heat required to raise the temperature of one mole of the gas by one degree.

So, if $\Delta$Q amount of heat goes to change the temperature of 'n' moles of a gas in a particular process, molar heat capacity 'C' can be mathematically given by:

C = $\frac{1}{n}\left[\frac{\Delta Q}{\Delta T}\right]$

In terms of differentials,

C = $\frac{1}{n}\frac{dQ}{dT}$

Two special cases are:-

(i) If volume is kept constant during the process then

C_V = $\frac{1}{n}{\left[\frac{\Delta Q}{\Delta T}\right]}_{V=constant}$

This is the molar heat capacity of the gas at constant volume

Note: Since $\Delta$U is independent of the process. $\Delta$U = n C_v $\Delta$T is true for all processes.

(ii) If pressure remains constant, then

C_p = $\frac{1}{n}{\left[\frac{\Delta Q}{\Delta T}\right]}_{p=constant}$

This is the molar heat capacity of the gas at constant pressure

Relation Between C_p and C_v

C_p - C_v = R

This is known as Mayer's relation.

The Values of C_p and C_v

If f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as

U = f/2 n RT

$\therefore$ $\Delta$U = f/2 n R$\Delta$T = n C_v$\Delta$T

$\therefore$ C_v = f/2 R

From Mayer's Relation

C_p = C_v + R

C_p = (f/2+1)R

And the ratio of specific heats

$\gamma$ = $\frac{C_p}{C_v}$ = $\frac{f/2+1}{f/2}$

$\gamma$ = $\frac{f+2}{f}$ = $1+\frac{2}{f}$

Example 1: Find the molar heat capacity of an ideal gas with adiabatic exponent '$\gamma$' for the polytropic process $P{V}^{\lambda }$= constant.

Solution: We have, from first law of thermodynamics

C = C_v + $\frac{PdV}{ndT}$ (n = number of moles)

We have, P V$\gamma$ = constant

From Ideal gas equation P V = n RT

Taking ratio, $V^{\lambda -1}$= Constant

Differentiating we get $\frac{dV}{dT}$ = - $\frac{V}{T(\lambda -1)}$

Putting it in the equation for 'C'.

C = C_v - $\frac{PV}{nT(\lambda -1)}$ = C_v - $\frac{nRT}{nT(\lambda -1)}$

= C_v - $\frac{R}{\lambda -1}$

C = $\frac{R}{\gamma -1}-\frac{R}{\lambda -1}$