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Questions
Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules.
Deduce an expression for molar specific heat of a monoatomic gas at constant volume.
Solution
Application of the law of equipartition energy in the specific heat of gas, Meyer’s relation CP − CV = R connects the two specific heats for one mole of an ideal gas.
The equipartition law of energy is used to calculate the value of CP – CV and the ratio between them γ = `"C"_"P"/"C"_"V"`. Here γ is called the adiabatic exponent.
(i) Monatomic molecule: Average kinetic energy of a molecule = `[3/2"kT"]`
Total energy of a mole of gas = `3/2"kT" xx "N"_"A" = 3/2"RT"`
For one mole, the molar specific heat at constant volume
CV = `"dU"/"dT" = "d"/"dT"[3/2"RT"]`
CV = `[3/2"R"]`
`"C"_"P" = "C"_"V" + "R" = 3/2"R" + "R" = 5/2"R"`
The ratio of specific heats, γ = `"C"_"P"/"C"_"V" = (5/2"R")/(3/2"R") = 5/3` = 1.67
(ii) Diatomic molecule: Average kinetic energy of a diatomic molecule at low temperature = `5/2"kT"`
Total energy of one mole of gas = `5/2"kT" xx "N"_"A" = 5/2"RT"`
(Here, the total energy is purely kinetic)
For one mole Specific heat at constant volume
CV = `"dU"/"dT" = [5/2"RT"] = 5/2"R"`
But, `"C"_"P" = "C"_"V" + "R" = 5/2"R" + "R" = 7/2"R"`
∴ γ = `"C"_"P"/"C"_"V" = (7/2"R")/(5/2"R") = 7/5` = 1.40
Energy of a diatomic molecule at high temperature is equal to `7/2"RT"`
CV = `"dU"/"dT" = "d"/"dT" [7/2"RT"] = 7/2"R"`
∴ CP = `"C"_"V" + "R" = 7/2"R" + "R"`
CP = `9/2"R"`
Note that the CV and CP are higher for diatomic molecules than the mono atomic molecules. It implies that increasing the temperature of diatomic gas molecules by 1°C requires more heat energy than monoatomic molecules.
∴ γ = `"C"_"P"/"C"_"V" = (9/2"R")/(7/2"R") = 9/7` = 1.28
(iii) Triatomic molecule:
(a) Linear molecule:
Energy of one mole = `7/2"kT" xx "N"_"A" = 7/2"RT"`
CV = `"dU"/"dT" = "d"/"dT"[7/2"RT"]`
CV = `7/2"R"`
CP = `"C"_"P"/"C"_"V" + "R" = 7/2"R" + "R" = 9/2"R"`
∴ γ = `"C"_"P"/"C"_"V" = (9/2"R")/(7/2"R") = 9/7` = 1.28
(b) Non-linear molecule:
Energy of a mole = `6/2"kT" xx "N"_"A" = 6/2"RT"` = 3RT
CV = `"dU"/"dT"` = 3R
CP = `"C"_"V" + "R"` = 3R + R = 4R
∴ γ = `"C"_"P"/"C"_"V" = (4"R")/(3"R") = 4/3` = 1.33
Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure.
The specific heat capacity varies with the temperature.
Notes
Students should refer to the answer according to their questions.
RELATED QUESTIONS
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Choose the correct answer from the option given below:
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