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Derive the ratio of two specific heat capacities of monatomic, diatomic and triatomic molecules. |
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Answer» Solution :Monoatomic molecule : Average kinetic energy of a molecule `=[3/2kT]` (i) TOTAL energy of a mole of gas `=3/2 kT xxN_A = 3/2 RT` (ii) For one mole , the MOLAR specific heat at constant volume `C_V = (DU)/(dT) =d/(dT) [3/2RT]` `C_V = [3/2R]` `C_P = C_V + R =3/2 R+R` `=5/2R` (III) The ratio of specific heats, `gamma=C_P/C_V=(5//2R)/(3//2R)=5/3=1.67` Diatomic molecule : Average kinetic energy of a diatomic molecule at low temperature `5/2kT`. (i) Total energy of one mole of gas `=5/2kT xxN_A = 5/2RT` (Here, the total energy is purely kinetic) (ii) For one mole specific heat at constant volume `C_V=(dU)/(dT)=[5/2RT]=5/2R` But, `C_P = C_V + R = 5/2R + R = 7/2R` `:. gamma=C_P/C_V=(7//2R)/(5//2R)=7/5=1.40` (iii) Energy of a diatomic molecule at high temperature is equal to `7/2` RT. `C_V=(dU)/(dT)=[7/2RT]=7/2R` `:. C_P = C_v + R = 7/2 R+R` `C_P=9/2R` It is noted that the `C_V and C_P` are higher for diatomic molecules than the monoatomic molecules. It is implied that to increase the temperature of diatomic gas molecules by `1^@C` it require more heat energy than monoatomic molecules. `:.gamma=C_P/C_V=(9//2R)/(7//2R)=9/7=1.28` Triatomic molecule : (i) Linear molecule : Energy of one mole `=7/2 kTxxN_A = 7/2RT` `C_V=(dU)/(dT)=d/(dT)[7/2RT]` `C_V=[7/2R]` `C_P=C_V+R=7/2R+R` `=9/2R` `gamma=C_P/C_V=(9//2R)/(7//2R)=9/7=1.28` (ii) Non - Linear molecule : Energy of one mole `=6/2kTxxN_A=6/2RT=3RT` `C_V=(dU)/(dT)=3R` `C_P=C_V+R` `=3R+R =4R` `gamma=C_P/C_V=(4R)/(3R)=4/3=1.33` |
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