posted by Anonymous .
A cubic box of volume 4.2 multiplied by 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?
My attempted solution:
.042^(1/3) = .3476 m for the area
.042^(2/3) = .1208 m for the area of a side
(273 + 195)/(273+20) = 1.597
1.597 * .1208 = .19292
.19292 was my answer.
Did I not take this far enough or is the entire set-up wrong?
Volume = 4.2*10^-2 m^3
Side length is the cube root of that:
a = 0.34760 m
Side area = a^2 = 0.1208 m^2
Absolute Temperature ratio = 1.5973
New atmospheric pressure
= 1.5973*1.013*10^5 N/m^2
= 1.618*10^5 N/m^2
Multiply that by the area of a side.
Your only mistake was not converting pressure from atmospheres to Pascals (N/m^2)