Using K E = 1 2 f n R T KE = \frac{1}{2}fnRT K E KE of O 2 O_{2} molecules = 1 2 5 n R T = \frac{1}{2}5nRT G i v e n ( K E ) 27 ∘ C = 2 ( K E ) T ∘ C Given\ (KE)_{27^{\circ}C} = 2(KE)_{T^{\circ}C} 5 2 n R ( 27 + 273 ) = 2 × 5 2 n R × T \frac{5}{2}nR(27 + 273) = 2 \times \frac{5}{2}nR \times T T = 600 K = 600 − 273 = 327 ∘ C T = 600\text{\ }K = 600 - 273 = {327\ }^{\circ}C

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27 °C is

Question 1:

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27 °C is

1227 °C

627 °C

327 °C

927 °C

Answer:

Solution:

Using $KE = \frac{1}{2}fnRT$

$KE$ of $O_{2}$ molecules $= \frac{1}{2}5nRT$

$Given\ (KE)_{27^{\circ}C} = 2(KE)_{T^{\circ}C}$

$\frac{5}{2}nR(27 + 273) = 2 \times \frac{5}{2}nR \times T$

$T = 600\text{\ }K = 600 - 273 = {327\ }^{\circ}C$

Prev Exam(s):

JEE MAIN

- 2023

, 8 Apr Shift-02

Chapter:

KINETIC THEORY OF GASES (KTG)

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