Question

A cubic box of volume \(6.15 \times {10^{ - 2}}\;{{\rm{m}}^{\rm{3}}}\) is filled with air at atmospheric pressure at 15°C. The box is closed and heated to 165°C. What is the net force on each side of the box?

Short answer

The force on each side of the box is 8207 N.

Step-by-step solution

Given data

The volume of the box is\(V = 6.15 \times {10^{ - 2}}\;{{\rm{m}}^{\rm{3}}}\).

The initial temperature is\({T_1} = {15^ \circ }{\rm{C}} = 288\;{\rm{K}}\).

The final temperature is\({T_2} = {165^ \circ }{\rm{C}} = 438\;{\rm{K}}\).

The initial pressure of the air inside the box is\({P_1} = 1\;{\rm{atm}}\).

Let\({P_2}\)be the final pressure of the air inside the box.

Let\(L\)be the sides of the cubic box.

Assume that there is no change in the volume of the box.

Concepts

The force on any side of the box equals the product of the pressure and area.

For this problem, you have to use\(\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}\)equation.

Calculation

Now, from the ideal gas law, you get:

\(\begin{aligned}{c}PV = nRT\\\frac{P}{T} &= \frac{{nR}}{V}\end{aligned}\)

The volume is constant in the process. Therefore,

\(\begin{aligned}{c}\frac{{{P_1}}}{{{T_1}}} &= \frac{{{P_2}}}{{{T_2}}}\\{P_2} &= {P_1}\frac{{{T_2}}}{{{T_1}}}\\{P_2} &= \left( {1\;{\rm{atm}}} \right)\frac{{438\;{\rm{K}}}}{{288\;{\rm{K}}}}\\{P_2} &= 1.52\;{\rm{atm}}\end{aligned}\)

Now, the extra pressure inside the box is:

\(\begin{aligned}{c}\Delta P &= {P_2} - {P_1}\\ &= 1.52\;{\rm{atm}} - 1\;{\rm{atm}}\\ &= 0.52\;{\rm{atm}}\\ &= 0.52 \times 1.013 \times {10^5}\;{\rm{Pa}}\end{aligned}\)

Again, you get:

\(\begin{aligned}{c}{L^3} &= V\\L &= {V^{1/3}}\\{L^2} &= {V^{2/3}}\\{L^2} &= {\left( {6.15 \times {{10}^{ - 2}}\;{{\rm{m}}^{\rm{3}}}} \right)^{2/3}}\end{aligned}\)

Therefore, the area of a side of the box is \({\left( {6.15 \times {{10}^{ - 2}}\;{{\rm{m}}^{\rm{3}}}} \right)^{2/3}}\).

Now, the force on the side of the box is:

\(\begin{aligned}{c}F &= {P_2}{L^2}\\ &= \left( {0.52 \times 1.013 \times {{10}^5}\;{\rm{Pa}}} \right) \times {\left( {6.15 \times {{10}^{ - 2}}\;{{\rm{m}}^{\rm{3}}}} \right)^{2/3}}\\ &= 8207\;{\rm{N}}\end{aligned}\)

Hence, the force on each side of the box is 8207 N.