Saved Bookmarks
| 1. |
The ratio of translational and rotational kinetic energies at `100K` temperature is `3:2`. Then the internal energy of one mole gas at that temperature is `(R=8.3j//"mol"-K]`A. `1175J`B. `1037.5J`C. `2075J`D. `4150J` |
|
Answer» Accordingly to law of equipartion of energy, energies equally distributed among its degree of freedom. Let transitional and rotaional degree of freedom be `f_(1)` and `f_(2)` `therefore (K_(T))/(K_(R))=(3)/(2)` and `K_(T)+K_(R)=U` Hence the ratio of transitional to rotational degrees of freedom is `3.2`. Since transiational degrees of freedom is `,`. the rotational degrees of freedom must be `2` `therefore` Internal energy `(U)=1xx(f_(1)+rf_(2))=(1)/(2)RT` `U=(1xx5xx8.3xx100)/(2)=U=2075J` |
|