Equal masses of methane and ethane have their total translational kinetic energy in the ratio 3: 1, then their temperatures are in the ratio is

Question

Equal masses of methane and ethane have their total translational kinetic energy in the ratio 3: 1, then their temperatures are in the ratio is (A) 5: 8 (B) 45: 8 (C) 15: 8 (D) 8: 5

Tardigrade Admin · Accepted Answer

E1=(3/2)× (M/16)RT1 E2=(3/2)× (M/30)RT2 ⇒ (E1/E2)=(30/16)(T1/T2) ⇒ (3/1)=(30/16)(T1/T2) ⇒ (T1/T2)=(8/5)

Q. Equal masses of methane and ethane have their total translational kinetic energy in the ratio 3 : 1, then their temperatures are in the ratio is

5: 8

45: 8

15: 8

8: 5

$E_{1} = \frac{3}{2} \times \frac{M}{16} R T_{1}$

$E_{2} = \frac{3}{2} \times \frac{M}{30} R T_{2}$

$\Rightarrow \frac{E _{1}}{E _{2}} = \frac{30}{16} \frac{T _{1}}{T _{2}}$

$\Rightarrow \frac{3}{1} = \frac{30}{16} \frac{T _{1}}{T _{2}}$ $\Rightarrow \frac{T _{1}}{T _{2}} = \frac{8}{5}$

Q. Equal masses of methane and ethane have their total translational kinetic energy in the ratio 3 : 1, then their temperatures are in the ratio is

5: 8

45: 8

15: 8

8: 5

E1​=23​×16M​RT1​ E2​=23​×30M​RT2​ ⇒E2​E1​​=1630​T2​T1​​ ⇒13​=1630​T2​T1​​ ⇒T2​T1​​=58​

$E_{1} = \frac{3}{2} \times \frac{M}{16} R T_{1}$

$E_{2} = \frac{3}{2} \times \frac{M}{30} R T_{2}$

$\Rightarrow \frac{E _{1}}{E _{2}} = \frac{30}{16} \frac{T _{1}}{T _{2}}$

$\Rightarrow \frac{3}{1} = \frac{30}{16} \frac{T _{1}}{T _{2}}$ $\Rightarrow \frac{T _{1}}{T _{2}} = \frac{8}{5}$