Monatomic gas: Using the law of equiportition of energy the average energy of molecule at temperature
\(T=\,\frac{3}{2}{{K}_{B}}T\)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 \({{C}_{V}}=\frac{du}{dT}=\frac{J}{2}\)
\({{C}_{P}}-{{C}_{V}}=R\Rightarrow {{C}_{P}}=\frac{7}{2}R\)\(r=\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{7}{5}\).
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