Question

A cubic box of volume 3.7 ? 10^-2 m³ is filled with air at atmospheric pressure at 20°C. The box is closed and heated to 195°C. What is the net force on each side of the box?

Answer 1

The expression for the volume of a cubic box of volume V and edge length h is V = h³.

The surface area of each face of the box is A = h²,

Therefore -

A = V^(2/3).

Now, according to Gay-Lussac's Law -

p1/T1 = p2/T2,

=> p2 = p1*T2/T1

So at 195°C (468 K).

p = (1 atm)*(468 K) / (293 K) = 1.60 atm = 161843 pascals = 161843 N/m³

This is the pressure on each of the 6 walls of the cube.

Now given that the volume of the cube is 3.7 * 10^(-2) m³,

So, area of each side is A = (3.7 * 10^(-2) m³)^(2/3) = 0.1098 m².

Pressure is force per unit area (F = p*A), so a pressure of 161843 N/m² exerted on a side with surface area of 0.1098 m². implies a force of (161843 N/m²)*(0.1098 m²) = 17774 N = 17.8 kN.

Therefore, net force on each side of the box = 17.8 kN Directed outward from the center of the box and normal to the sides of the box.