Chapter 13, Problem 44P

To determine

To Calculate: The net force on each side of the box.

Answer to Problem 44P

The net force on each side of the box is $7.7\times {10}^2\text{N}$ .

Explanation of Solution

Given info:

The volume of the box is, $V=5.1\times {10}^{-2}{\text{m}}^{\text{3}}$ .

The temperatures outside the box, $T_1=20°\text{C}$

The temperatures inside the box $T_2=180°\text{C}$ .

Outside pressure, $P_1=1\text{atm}$ .

Formula used:

Write the expression for the ratio of pressure and temperature in case of ideal gas.

  $\frac{P_2}{T_2}=\frac{P_1}{T_1}$

Here, $P_2$ is the inside temperature,

  $P_1$ is the outside pressure,

  $T_2$ is the inside temperature,

  $T_1$ is the outside temperature.

Area of the square, $A_s=\text{side}\text{×}\text{side}$

The net force is calculated as,

  $F=\Delta P\times A$

Here, $\Delta P$ is the change in pressure,

  $A$ is the Area of square box.

Calculation:

The net force on each side of the box is equals to the product of the area of the box and the pressure difference between inside and outside box.

The outside pressure is 1 atm while the inside pressure is calculated with the help of ideal gas law such that,

  $\begin{array}{l}\frac{P_2}{T_2}=\frac{P_1}{T_1}\\ \Rightarrow P_2=\frac{T_2}{T_1}{P}_1=\left(1\text{atm}\right)\left(\frac{273+180}{273+20}\right)\text{K}\\ P_2=1.55\text{atm}\end{array}$

The area of the square box is,

  $\begin{array}{l}{A}_s=\text{side×side}={\left(L\right)}^2\\ A_s={\left({\left(\text{volume}\right)}^{\text{1/3}}\right)}^2={\left(5.1\times {10}^{-2}{\text{m}}^{\text{3}}\right)}^{2/3}\\ A_s=0.14{\text{m}}^{\text{2}}\end{array}$

The net force on the side of the box is,

  $\begin{array}{c}F=\Delta P\times A_s\\ =\left(P_2-P_1\right)\times A_s\\ =\left(1.55\text{atm}-1\text{atm}\right)\times 0.14{\text{m}}^{\text{2}}\\ =7.7\times {10}^2\text{N}\end{array}$

Conclusion:

The net force is $7.7\times {10}^2\text{N}$ .