# Force on each wall of a box

• jennypear
In summary, a cubic metal box with sides 32.0 cm contains air at a pressure of 1.00 atm and a temperature of 294 K. When the box is sealed and heated to a temperature of 396K, the pressure inside the box increases to 1.35 atm. The force on each wall of the box due to this pressure difference is 0.03584 N. It is important to use consistent SI units in calculations to obtain accurate results.

#### jennypear

A cubic metal box with sides 32.0 cm contains air at a pressure of 1.00 atm and a temperature of 294 K. The box is sealed so that the volume is constant and it is heated to a temperature of 396K. Find the force on each wall of the box due to the increased pressure within the box. [The outside air is at 1 atm of pressure.]

I started out PV=nRT

volume of the box = .32^3=0.328m^3*(1L/1x10^-3)=32.8L
n=PV/RT
1atm(32.8L)/(.08207 atm L/mol*K)(294K)=1.36mol

P=nRT/V
[1.36mol(.08207 atm L/mol*K)396K]/32.8L
=1.35 atm

F=P*A
=1.35 atm*.32^2=.138

There is no need to calculate n, since it is constant, but it is not wrong to do so. Assuming your calculation of the pressure is correct, it looks like you have just neglected to account for the 1atm pressure still on the outside.

how do you account for that?

jennypear said:
how do you account for that?

The pressure on the outside is still 1 atm. It is the pressure difference between inside and outside that produces the net force on the walls.

OlderDan said:
The pressure on the outside is still 1 atm. It is the pressure difference between inside and outside that produces the net force on the walls.

so that would be
F=deltaP*A
.35*.32^2=.03584

that came up as incorrect

jennypear said:
so that would be
F=deltaP*A
.35*.32^2=.03584

that came up as incorrect

What have you done about units? What are the units of the answer you are trying to find? How do you express one atmosphere in those units?

Your whole numerical computations need to be redone,as they're wrong...The volume of the box,for example

$$V_{\mbox{box}}=(32\cdot 10^{-2}\mbox{m})^{3}=2^{15}\cdot 10^{-6}\mbox{m}^{3}=32.784\cdot 10^{-6}\mbox{m}^{3}\simeq 32.8\cdot 10^{-3}\mbox{m}^{3}$$

Use SI-mKgs units...

Daniel.

Like Dex said...units!

Since you are working in SI units, your volume must be in meters and your pressure must be in Pa (n/m^2). Only then will your answer be in Newtons.

## 1. What is the force on each wall of a box?

The force on each wall of a box is the amount of pressure or push that is exerted on the walls by the contents of the box. This force is typically measured in Newtons.

## 2. How is the force on each wall of a box calculated?

The force on each wall of a box can be calculated by multiplying the mass of the contents of the box by the acceleration due to gravity. This is known as the weight of the box, and it is equal to the force exerted on each wall.

## 3. Does the size of the box affect the force on each wall?

Yes, the size of the box can affect the force on each wall. The larger the box, the more weight it can hold, and therefore the greater the force on each wall.

## 4. How does the shape of the box impact the force on each wall?

The shape of the box can also have an impact on the force on each wall. Boxes with a wider base or a longer length will distribute the weight more evenly, resulting in less force on each wall compared to boxes with a smaller base or shorter length.

## 5. Can the force on each wall of a box change?

Yes, the force on each wall of a box can change. This can happen if the contents of the box are shifted or if the box is subjected to external forces such as wind or vibrations. Additionally, as the contents of the box are removed or added, the force on each wall will change accordingly.