Question

36, The kinetic energy of 1 gram mole of a gas at normal
temperature and pressure is (R = 8.31 l/mole-K)
(a) 0.56 X 10') (b) 1.3 x 10'
(c) 2.7 X 10'
d) 3.4 x 10'​

Answer 1

Answer:

K.E . = 3.4 x 10³  J

Option d. is correct.

Explanation:

Given :

Normal temperature i.e. 273 K

Value of R = 8.31  J mol⁻¹ K⁻¹

We have to find K.E.

We have formula for K.E.

K.E. =  3 / 2 × R T

Putting values here :

K.E.  =  3 / 2 × 8.31 × 273  kJ

K.E =  1.5 × 8.31 × 273  kJ

K.E. =  3,402.9 kJ

K.E . = 3.4 x 10³  J

Hence we get answer.