Question

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 1

Answer:3.92475J

Step-by-step 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

Answer 2

9.57 × 10⁷ J

Step-by-step 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