A 1.0-mol sample of hydrogen gas has a temperature of 27◦C. (a) What is the total kinetic energy of all the gas molecules in the sample? (b) How fast would a 65-kg person have to run to have the same kinetic energy?

(a) K. E. = 3,741.512‬ J, (b) v = v = 10.73 m/s

Explanation:

a) Ideal Gas Formula in term of K. E.

K. E. = 3/2 n R T = 3/2 n R T (R is universal gas constant = 8.314472 J/mol

K. E. = (3/2) * 1.0 mol * 8.314472J/mol * 300 K (t = 27° C = 300 K)

K. E. = 3,741.512‬ J

b) K. E. = 1 / 2 m v²

⇒ v = spuare root (2 K. E. / m)

v = square root (2 * 3,741.512‬ K / 65 kg)

v = 10.73 m/s