Question

7.9mol of helium are in a 14L cylinder. The pressure gauge on the cylinder reads 67psi . What are

(a) the temperature of the gas in ?C and

(b) the average kinetic energy of a helium atom?

Answer 1

given that ::

number of moles, n = 7.9 mol

volume of the cylinder, V = 14 L = 0.014 m³

pressure, p = 67 psi

psi change into Pa :

(67 + 14.7)psi x (1 atm / 14.7 psi) x (1.01 x 10⁵ Pa/atm) = 5.557 x 10⁵ Pa

(a) the temperature of the gas in ⁰C is given as ::

using an ideal gas equation,

pV = nRT                                { eq. 1 }

or T = pV / nR

where, R = gas constant = 8.314 J/mol-K

inserting the values in above eq.

T = (5.557 x 10⁵ Pa) (0.014 m³) / (7.9 mol) (8.314 J/mol-K)

T = (7779.8) / (65.68) K

T = 118.4 K

kelvin change into degree celsius, (118.4 - 273) = -154.6 degree celsius

T = -154.6 ⁰C

(b) the average kinetic energy of a helium atom is given as ::

K.E_avg = (3/2) kT                          { eq. 2 }

where, k = boltzmann constant = 1.381 x 10^-23 J/mol-K

inserting the values in eq.2,

K.E_avg = (3/2) (1.381 x 10^-23 J/mol-K) (118.4 K)

K.E_avg = 245.26 x 10^-23 J

or   K.E_avg = 2.45 x 10^-21 J