The ratio of translational to rotational kinetic energies at 100 K temperature is 3: 2. The internal energy (in J) of one mole gas at that temperature is (R=8.3 J mol -1 K -1)

Question

The ratio of translational to rotational kinetic energies at 100 K temperature is 3: 2. The internal energy (in J) of one mole gas at that temperature is (R=8.3 J mol -1 K -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

According to law of equipartition of energy, energies equally distributes among its degree of freedom.
Let translational and rotational degree of freedom be

f_{1}

and

f_{2}

, respectively.

∴ \frac{K _{T}}{K _{R}} = \frac{3}{2}

and

K_{T} + K_{R} = U

Hence, the ratio of translational to rotational degrees of freedom is

3 : 2

. Since translational degrees of freedom is

3

, the rotational degrees of freedom must be

2

.

∴

Internal energy

(U) = 1 \times (f_{1} + f_{2}) \times \frac{1}{2} RT

U = \frac{1 \times 5 \times 8.3 \times 100}{2}

\Rightarrow U = 2075 J

Q. The ratio of translational to rotational kinetic energies at 100K temperature is 3:2. The internal energy (in J) of one mole gas at that temperature is (R=8.3Jmol−1K−1)

Q. The ratio of translational to rotational kinetic energies at $100 K$ temperature is $3 : 2$ . The internal energy (in $J$ ) of one mole gas at that temperature is $(R = 8.3 J m o l^{- 1} K^{- 1})$