InterviewSolution
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InterviewSolution
| 1. |
Derive Eq. (6.4c), making use of the Boltzmann distribution. From Eq.(6.4c) obtain the expression for molar vibration heat capacity `C_("V vib")` of diatomic gas. Calculate `C_("V vib")` for `cl_(2)` gas at the temperature `300K`. The natural vibration frequency of these molecules is equal to `1,064.10^(14)s^(-1)`. |
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Answer» by defination `lt E gt =(Sigma ^(E_(V)e^(-E_(v)//kT)))/(Sigma exp(-E_(v)//kT)) = ((del)/(delbeta)Sigma_(v=0)^(oo)e^(betaE_(v)))/(Sigma_(v=0)^(oo)e^(-betaE_(v)))` `=-(del)/(del beta) In Sigma_(v=0)^(oo)e^(-beta(v+1//2)ħomega), beta=(1)/(KT)` `=-(del)/(delbeta) In e^(-1//2beta ħ omega)(1)/(1-e^(-beta ħomega))` `= (del)/(del beta)[-(1)/(2) ħomegabeta-In(1-e^(-beta ħ onrga))]` `(1)/(2) ħ omega+( ħ omega)/(e^( ħ oega//kT)-1)` Thus for one gm mole of diatomic gas `C_("v vib")=N(dellt E gt)/(delT)=(R((ħ omega)/(KT))^(2)e^(ħ omega//kT))/((e^(ħ omega//kT)-1)^(2))` where `R=Nk` is the gas constant. In the present case `(ħ omega)/(KT)= 2.7088` and `C_("V 1 vib")= 0.56R` |
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