Specific heat capacity

Monatomic gas: Using the law of equiportition of energy the average energy of molecule at temperature


Total internal energy of a gas is \(u=\frac{3}{2}{{K}_{B}}T\times {{N}_{A}}=\frac{3}{2}RT\)

Cv [Monatomic gas] = \(\frac{du}{dt}=\frac{3}{2}RT\)

For an ideal gas.

CP – CV = R

CP = J/2 R

Ratio of specific heats\(r=\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{\frac{J}{2}}{\frac{3}{2}}=\frac{J}{3}\).

Diatomic gases: Total internal energy [law of equiportition of energy]

\(u=\frac{5}{2}{{K}_{B}}T\times {{N}_{A}}=\frac{5}{2}RT\).

Then \({{C}_{V}}=\frac{du}{dT}=\frac{J}{2}\)

\({{C}_{P}}-{{C}_{V}}=R\Rightarrow {{C}_{P}}=\frac{7}{2}R\)


Polyatomic gases: We can write the total internal energy \(U=\left[ \frac{3}{2}{{K}_{B}}T+\frac{3}{2}{{K}_{B}}T+F{{K}_{B}}T \right]{{N}_{A}}\).

i.e., \({{C}_{V}}=\frac{du}{dT}=\left( 3+f \right)R\).

Cp = (4 + f) R.

Where f = no. of vibrational modes \(r=\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{\left( 4+f \right)}{\left( 3+f \right)}\)

Specific heat capacity of solids

Consider a solid of “N” atoms, each are vibrating about its mean position.

Average energy = 2 x ½ KBT [one – dimension]

Average in three dimension = 3KBT

u = n x 3KBT x NA At solids DY is negligible

So DQ = DU.

\(C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{\Delta T}=3R\).

Specific heat capacity of heater:

u = n x 3KBT x NA = 3nRT

n = Number of atoms in molecules

c = \(\frac{\Delta q}{\Delta T}\)= 3nR

Example: The average KE and the rms velocity of the molecule in a sample of oxygen gas at 300k are 6.21 x 10⁻²¹  and 484m/s respectively. If the temperature is doubled find the value new KE and rms speed.

a) KE = 12.42 x 10⁻²²rms = 685 m/s

b) KE = 11.42 x 10⁻²²rms = 686.5 m/s

c) KE = 11.42 x 10⁻²³rms = 684.4 m/s

d) None of these

Answer: d)

Solution: \(E\,=\frac{3}{2}KT$ rms, speed $c\,=\sqrt{\frac{3RT}{M}}\).

On doubling temperature KE = 2E and rms speed = √2c