What is the total translational kinetic energy of the air in an empty room that has dimensions 9.00m*14.0m*5.00m if the air is treated as an ideal gas at 1.00 atm?

Answer:3.92475J

Explanation:

ideal gas equation

PV=nrT

P=pressure

V=volume 9*5*14=630m^3

n=mole

Mole=mass/molecular mass

molecular mass of air=28.1g/mol

R=gas constant

T=temperature in kelvin

1.01*10^5*630 = (m/28) * 8.314j/molK * 273K

784959.5677356082g

784.95kg

Kinetic energy=0.5*m*v^2

Velocity of air

0.5*784.95kg*0.1^2

K. E=3.92475J

6

9.57 * 10⁷ J

Explanation:

The translational kinetic energy of an ideal gas is K. E = 3/2nRT. But from ideal gas law, PV = nRT, so

K. E = 3/2PV where pressure of air = 1.00 atm = 1.013 * 10⁵ Pa and V = volume of room = 9.00m * 14.0m * 5.00m = 630 m³

K. E = 3/2PV = 3/2 * 1.013 * 10⁵ Pa * 630 m³ = 9.57 * 10⁷ J