Question

A cubic box of volume 5.1 x 10 -2m3 is

Answer 1

Answer:

21544 N

Explanation:

Let the atmospheric pressure be 101325 Pa

The side length of the cubic box:

$V = d^3 = 0.051m^3$

$d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m$

The area of the cubic box:

$A = d^2 = 0.371^2 = 0.1375 m^2$

20 C = 20 + 273 = 293 K

180 C = 180 + 273 = 453 K

As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up

$P_2 = P_1\frac{T_2}{T_1} = 101325\frac{453}{293} = 156656 Pa$

The net force on each side of the box it its pressure times side area

$F = P_2A = 156656 * 0.1375 = 21544 N$