The total kinetic energy of 1 mole of oxyge 27^{\circ} {\C} is : [Use universal gas constant ( {\R})=8.31 {\J} \/ {\mole} {\K} [27-Jan-2024 Shift 2]

JEE Main 27-Jan-2024 Shift 2 Solved Paper

Section: Physics

Question : 40 of 90

Marks: +1, -0

The total kinetic energy of 1 mole of oxyge 27∘C is :
[Use universal gas constant (R)=8.31J∕mole‌K

[27-Jan-2024 Shift 2]

Go to Question:

Prev Question

Next Question