Question 646: Thermodynamics & KTG

Question

The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at $57^{\circ} \mathrm{C}$ is (molecular masses of oxygen and helium are 32 and 4 respectively.)

Accepted Answer

Step 1: Recall the formula for root mean square (r.m.s.) speed
$ v_{	ext{rms}} = \sqrt{\frac{3kT}{m}} $
where $k$ = Boltzmann constant, $T$ = absolute temperature, $m$ = mass of one molecule.
Equivalently, with molar mass $M$:
$ v_{	ext{rms}} = \sqrt{\frac{3RT}{M}} $
where $R$ = universal gas constant and $M$ is molar mass.
Step 2: Equating r.m.s. speeds
For helium at temperature $T_	ext{He}$ and oxygen at $T_	ext{O2}$:
$ \sqrt{\frac{3RT_{	ext{He}}}{M_{	ext{He}}}} = \sqrt{\frac{3RT_{	ext{O2}}}{M_{	ext{O2}}}} $
Squaring and simplifying:
$ \frac{T_	ext{He}}{M_	ext{He}} = \frac{T_	ext{O2}}{M_	ext{O2}} $
$ T_	ext{O2} = T_	ext{He} \cdot \frac{M_	ext{O2}}{M_	ext{He}} $
Step 3: Substitute values
$T_	ext{He} = 57^\circ 	ext{C} = 57 + 273 = 330\ 	ext{K}$.
$M_	ext{He} = 4$, $M_	ext{O2} = 32$.
$ T_	ext{O2} = 330 \cdot \frac{32}{4} = 330 \cdot 8 = 2640\ 	ext{K} $
Ã¢ÂÂ Final Answer:
$ \boxed{2640\ 	ext{K}} $
Correct Option:
C
.

Answer

1320 K

Answer

2240 K

Answer

2640 K

Answer

3230 K

Thermodynamics & KTG

The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at 57^{\circ} \mathrm{C} is (molecular masses of oxygen and helium are 32 and 4 respectively.)

Choose the correct answer: