Saved Bookmarks
| 1. |
Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules. |
|
Answer» Solution :Application of law of equipartition energy in specific heat of a gas : Meyer's relation `C_(P)-C_(Y)=R` connects the two specific heats for one mole of an ideal gas. Equipartition law of energy is used to calculate the value of `C_(P)-C_(V)` and the ratio between them `gamma=(C_(P))/(C_(V))` Here `gamma` is called adiabatic exponent. (i) Monotomic molecule: Average kinetic energy of a molecule `=[(3)/(2)kT]` Total energy of a mole of gas `(3)/(2)kTxxN_(A)=(3)/(2)RT` For one mole, the molar specific heat at constant volume `C_(V)=(dU)/(dT)=(d)/(dT)[(3)/(2)RT]` `C_(V)=[(3)/(2)R]` `C_(P)=C_(V)+R=(3)/(2)R+R=(5)/(2)R` The ratio of specific heats, `gamma=(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)kTxxN_(A)=(5)/(2)RT` (Here, the total energy is PURELY kinetic) For one mole Specific heat at constant volume `C_(V)=(dU)/(dT)=[(5)/(2)RT]=(5)/(2)R` But `C_(P)=C_(V)+R=(5)/(2)R+R=(7)/(2)R` `THEREFORE gamma=(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 `C_(V)=(dU)/(dT)=[(7)/(2)RT]=(7)/(2)R` `therefore C_(P)=C_(V)+R=(7)/(2)R+R` `C_(P)=(9)/(2)R` Note that the `C_(V)` and `C_(P)` are higher for diatomic molecules than the monoatomic molecules. It implies that to increase the temperature of diatomic gas molecules by `1^(@)C` it require more heat energy than monoatomic molecules. `therefore gamma=(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)kTxxN_(A)=(7)/(2)RT` `C_(V)=(dU)/(dT)=(d)/(dT)[(7)/(2)RT],C_(V)=(7)/(2)R` `C_(P)=C_(V)+R=(7)/(2)R+R=(9R)/(2)` `therefore gamma=(C_(P))/(C_(V))=((9)/(2)R)/((7)/(2)R)=(9)/(7)=1.28` (b) Non-linear molecule Energy of a mole `=(6)/(2)kTxxN_(A)=(6)/(2)RT=3RT` `C_(V)=(dU)/(dT)=3R` `C_(P)=C_(V)+R=3R+R=4R` `therefore gamma=(C_(P))/(C_(V))=(4R)/(3R)=(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. |
|