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

the ima (higher than 1m) will be greater than the ima with the height of 1m.