Question

Three moles of ideal gas are in a rigid cubical box with sides of length 0.470 m .

What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 16.0 ?C?

What is the force when the temperature of the gas is increased to 105 ?C?

Answer 1

a)
Using the ideal gas equation, PV = nRT,
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
V = (0.470)³ = 0.103823 m³.
T = 16 + 273.15 = 289.15 K
P = nRT/V
= [3 x 8.314 x 289.15] / 0.103823
= 69464.18 Pa

Force exerted on the wall = Pressure x Area
= 69464.18 x 6 x (0.47)²
= 92067.82 N

b)
T = 105 + 273.15 = 378.15 K
P = nRT/V
= [3 x 8.314 x 378.15] / 0.103823
= 90845.16 Pa

Force exerted on the wall = Pressure x Area
= 90845.16 x 6 x (0.47)²
= 120406.18 N