Question

Define Cp and Cv,derive an expression for the ratio of Cp/Cv for a diatomic gas

Answer 1

Thermodynamics :D

Here, Symbols have usual meanings in Thermodynamics

1)$C_{v}$ (Molar Heat Capacity at constant Volume)

It is defined as Heat given per mole of the gas per unit rise in temperature at constant volume.

$C_v = (\frac{\Delta Q}{n \Delta T } )_{constant\:\:volume}$

$C_p$ (Molar Heat Capacity at constant Pressure)

It is defined as Heat given per mole of the gas per unit rise in temperature at constant Pressure.

$C_p = (\frac{\Delta Q}{n \Delta T } )_{constant\:\:Pressure}$

2)Diatomic Gas (At temperature 200K < T < 5000K to ignore vibrational effects)
=>At mentioned Temperature , Diatomic gas posses energy in the form of Kinetic and rotational motion.
In Diatomic gases , the molecules are assumed to be in the shape of In Diatomic gases , the molecules are assumed to be in the shape of dum  bells, two hard spheres of negligible size at a separation.

=> In 3-D co-ordinate system, Diatomic gas will have 3 independent terms/ways to posses energy in the form of kinetic.

As:

$E_{kinetic}=\frac{1}{2}mv_x^{2}+\frac{1}{2}mv_y^{2}+\frac{1}{2}mv_z^{2}$

=>And also, it will have 2 independent ways to posses energy in the form of rotational motion.

$E_{rotaional}=\frac{1}{2}I_x\omega_x^{2}+\frac{1}{2}I_y\omega_y^{2}$

Term like : $\frac{1}{2}I_z\omega_z^{2}$ is dropped as size of atom is neglizible ,i.e. $I_z$ is negligible when compared to $I_x, I_y$ .

∴ The no. of independent terms to posses energy are (3+2)=5 .

3) Here,Comes a new Term :

The number of independent ways /terms to posses energy of a system/particle/molecule is called its Degree of Freedom.

For Diatomic gases, the degree of Freedom is 5 at mentioned temperature.

According to Equipartition Theorem, the average energy of a molecule in a Diatomic gas is $\frac{5}{2} kT$ .

4) Now, Consider a sample of amount 'n' moles of Diatomic gas.

The total no. of molecules is $nN_A$ where $N_A$ is Avogadro Number.

Then, the Internal Energy of a diatomic gas is given by :

$U=nN_A(\frac{5}{2}kT)= \frac{5}{2}nRT$

The Molar Heat Capacity at constant Volume is

$C_v=\frac{dU}{ndT} =\frac{5R}{2}$

The molar heat capacity at constant Pressure:

$C_p=C_v +R =\frac{5R}{2}+R=\frac{7R}{2}$

$\gamma=\frac{C_p}{C_v} = \frac{\frac{7R}{2}}{\frac{5R}{2}}=\frac{7}{5} =1.40$

Hence,For a Diatomic Gas value of $\gamma= 1.40$.

Answer 2

Answer:

good answer,but in a difficult way