Degree of Freedom
Trending Questions
At room temperature, the number of degrees of freedom of linear diatomic molecule is
3
4
5
6
- 3R2
- 5R2
- 4R
- 2R
If the degree of freedom of a gas are f, then the ratio of two specific heats CpCv is given by
2f+1
1−2f
1+1f
1−1f
- 2f+1
- 1−2f
- 1+1f
- 1−1f
- Diatomic
- Mixture of diatomic and polyatomic molecules
- Monoatomic
- Polyotomic
At room temperature, the number of degrees of freedom of linear diatomic molecule is
3
4
5
6
We know that, for an ideal monoatomic gas, the total internal energy is a result of the translational kinetic energy only. For the oxygen molecule O2 (assume it doesn't oscillate along the bond axis), which rotates as well as translates, choose the correct statement(s) from the following
Change in translational kinetic energy of a sample of n moles = nCVΔT
Translational kinetic energy < rotational kinetic energy
- Translational kinetic energy > rotational kinetic energy
- Translational kinetic energy = rotational kinetic energy.
We know that, for an ideal monoatomic gas, the total internal energy is a result of the translational kinetic energy only. For the oxygen molecule O2 (assume it doesn't oscillate along the bond axis), which rotates as well as translates, choose the correct statement(s) from the following
Change in translational kinetic energy of a sample of n moles = nCVΔT
Translational kinetic energy < rotational kinetic energy
- Translational kinetic energy > rotational kinetic energy
- Translational kinetic energy = rotational kinetic energy.
- Q=2RT
- Q=RT
- Q=3RT
- Q=4RT
Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy in this process. If n1 and n2 are the respective number of molecules of the gases, the temperature of the mixture will be
n1T1+n2T2n1+n2
n2T1+n1T2n1+n2
T1+n2n1T2
T2+n1n2T1
- 3
- 75
- 53
- 32
- 52R
- 23R
- R
- R3
- C=∞
- C=0
- C=R
- C=2R
- C=∞
- C=0
- C=R
- C=2R
(Take (5.66)0.4=2)
- 2
- 3
- 5
- 6
- Diatomic
- Mixture of diatomic and polyatomic molecules
- Monoatomic
- Polyotomic