Question

Relation between specific heat and degree of freedom.

Answer 1

Step 1: Formula used

1. $\gamma =\frac{C_P}{C_V}$ where $\gamma$ is the ratio of specific heat $C_P$​ is the specific heat at constant pressure and$C_V$ is the specific heat at constant volume
2. $U=\frac{f}{2}{K}_{B}T$ where $U$ is the internal energy, $f$ is the degree of freedom $K_B$ is Boltzmann's constant $T$ is the temperature

Step 2: Finding the relation between specific heat and degree of freedom

The ratio of the specific heat $\gamma$ is given by

$\gamma =\frac{C_P}{C_V}$ ………(i)

The equation for internal energy with degree of freedom ′$f$′ is as follows:

$U=\frac{f}{2}{K}_{B}T$ …..(ii)

We know that total energy is

$E=U\times N$ …..(iii)

$N$ being the number of molecules
Adding the value from the equation now (iii)

$E=\frac{f}{2}N{K}_{B}T$

Take $R=N{K}_B$ and we have

$\Rightarrow E=\frac{f}{2}RT$……(iv)

The specific heat at constant volume is currently understood to be provided by,

$C_V={\left(\frac{\partial E}{\partial T}\right)}_V$

Equation (iv) value's for $E$ is entered, and after further solving, we have

$\Rightarrow C_V=\frac{f}{2}R$

Now, the formula for the relationship between the specific temperatures is,

$C_P-C_V=R$

On putting the value of $C_V$ n the above equation from equation (v), we get

$C_P=\frac{R}{2}\left(f+2\right)$

Input the values from equations (v) and (vi) now (i),

$$\begin{gathered} \gamma =\frac{{\displaystyle \frac{R}{2}}\left(f+2\right)}{{\displaystyle \frac{R}{2}}fT} \\ \Rightarrow \gamma =1+\frac{2}{f} \\ \Rightarrow f=\frac{2}{\gamma -1} \end{gathered}$$

Hence, this is the relationship between the ratio of specific heats$\left(\gamma \right)$ of gas and degree of freedom $"f"$.