##
__Heat Capacity Of Gases:__

__Heat Capacity Of Gases:__

Heat capacity of a substance is defined as the amount of heat required to rise of the temperature by one degree. Heat capacity per gram of substance is called specific heat and per mole called molar heat capacity.

Thus,

__Molar heat capacity__=__Molar mass__×__Specific heat__**Cp = M × Cp**

**Cv = M × Cv**

Where

**Cp**and**Cv**are the molar heat capacities at constant pressure and constant volume respectively.**Cp**and**Cv**are their specific heats.

__Problem:__
The specific heat at constant pressure and constant volume are 0.125 and 0.075 Cal gm⁻¹K⁻¹ respectively. Calculate the molecular weight and atomicity of the gas. Name the gas if possible.

__Answer :__M = 40 and ⋎ = 1.66(mono-atomic), Ar(Argon).

__constant volume__and

__constant pressure__.

These are Represented as,

**Cv**= (dq/dT)v = (dU/dT)v

**Cp**= (dq/dT)p = (dU/dT)p + P(dV/dT)p

__Difference in Heat Capacities of an Ideal Gas:__
If the gas is assumed to be ideal, then,

PV = nRT and (dU/dT)v = (dU/dT)p

Again, P(dV/dT)p = nR

Thus for an ideal gas, PV = nRT; or, P(dV/dT)p = nR.

Again,

**Cp**= (dU/dT)p + P(dV/dT)p
or, Cp = Cv + {P×(nR/P)}

or,

**Cp = Cv +nR**
For 1 mole ideal gas

**Cp = Cv +R**

__Molar Heat capacities of Gases:__

From the above two descriptions, it is clear that

**Cp**and**Cv**. Since for**Cp**, some mechanical work is required as additional energy to absorbed for definite piston from volume**V₁**to**V₂**. Thus , Cp - Cv = Mechanical Work = PdV = P(V₂-V₁) = PV₂-PV₁ = R(T+1) - RT = R
Again, Cp = Cv+R for 1 mole ideal gas.

Now let us find the expression of Cv from the point of view of Kinetic Theory.

Cv = Energy required to increase transnational kinetic energy of 1 mole gas for rise of 1° temperature + energy required to increase inter molecular energy of 1 mole gas for rise temperature of 1°.

Increase of transnational K.E. = (3/2)R(T+1) - (3/2)R = (3/2) R for 1 mole gas for 1° rise in temperature.

__Mono-atomic Gases:__Cv = (dU/dT)v = (3/2)R and Cp = (5/2)R.

Thus Î³ = Cp/Cv = 5/3 ≃ 1.667

__Poly-atomic Gases:__Linear → Cv = (dU/dT)v = (3/2)R + R +(3N - 5)R

Non-linear → Cv = (dU/dT)v = (3/2)R + (3/2)R +(3N - 6)R

Where N is the number of particles.

__Molar Heat Capacities for Diatomic Molecules:__For diatomic molecule N = 2.

Thus Cv = (3/2)R + R + R = (7/2)R and Cp = (9/2)R.

∴ ⋎ = Cp/Cv = 9/7 ≃ 1.286

__Molar Heat Capacities for Tri-atomic Molecules:__For tri-atomic molecule N = 3.

Thus Cv(Linear) =