# Homework Help: Concerning Theory of Gases

1. Oct 24, 2009

### pc2-brazil

Good morning,

(This is not homework; just curiosity)
I found a little problem concerning Theory of Gases. I was able to solve it, but I want confirmation about if this is correct or not.
1. The problem statement, all variables and given/known data
Suppose I have a cube-shaped box, with edges measuring 20 cm = 0,2 m, which contains air, with pressure of 1 atm = 101325 Pa, in the temperature of 300 K. Suppose the volume of the box remains constant, and, then, I raise the temperature up to 400 K.
I want to find the net force acting upon each face of the box.

2. Relevant equations
As the volume of the box remains constant, I will use the equation P1/T1 = P2/T2, in which P and T are, respectively, pressure and temperature.
Other equation: P = F/A (force = pressure/area), with force in Newtons, pressure in Pascals and area in m².

3. The attempt at a solution
Firstly, I will find the pressure (in Pascals) inside the box at 400 K using the first equation above:
$$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$$
initial pressure is 101325 Pa, initial temperature is 300 K and final temperature is 400 K.
$$\frac{101325}{300}=\frac{P_{2}}{400}$$
therefore, the final pressure P2 = 135100 Pa.
Now, I will find the force acting upon each face. The area of each face is the square of the edge:
A = 0,22
A = 0,04 m²
P = F/A
F = PA
F = 135100.0,04
F = 5404 N = 5.4 kN
Is this correct?

Thank you in advance.

Last edited: Oct 24, 2009
2. Oct 24, 2009

### D H

Staff Emeritus
You got it.

3. Oct 24, 2009

### Andrew Mason

Not exactly. The question asks for the NET force. What other force (pressure) is acting on the sides of the boxes?

AM

4. Oct 24, 2009

### pc2-brazil

I was forgetting the fact that the side of the box never moves because of the force exerted by the gas. Then, 5,4 kN is the force exerted by the gas, not the net force.
If the side of the box remains at rest, then its net force should be zero, because the side is pushed by the force exerted by the gas, but this force is balanced by the friction exerted on it by the adjacent sides: the roof, the floor and the other two sides. Is this right?

5. Oct 24, 2009

### Andrew Mason

That balances the net force but that is not the force you have found. Put it this way: Is there a net force on the box sides when you begin? If the sides were made out of light cardboard taped with scotch tape, would it explode just sitting there? Using your calculation, it would (it would have about 4 kN of force on each side.)

AM

6. Oct 24, 2009

### pc2-brazil

I think that I'm not understanding correctly what you are saying. When I begin (before the temperature change), the net force on a side of the box is zero. After the temperature change, it is also zero, isn't it? It means that the sum of all contact forces of this side of the box with the adjacent sides balances the 5.4 kN exerted by the gas. Thus, before the temperature change, the sum of these contact forces was 101325 Pa * 0,04 m² = 4053 N.

7. Oct 24, 2009

### Andrew Mason

It is sitting in air with pressure 1 atm and temperature 300k - at least that is what appears from the question. It does not say it is sitting in a vacuum. The box has zero net outward force on each side to begin and puts no outward strain on the box structure. Why is that? When the temperature of the air in the box is heated there is a net outward force. This force has to be balanced by the box structure, but that is not what the question is asking. It is asking what that net (outward) force is.

AM

Last edited: Oct 24, 2009
8. Oct 24, 2009

### pc2-brazil

I think I got you now.
Inside the box, at 300 K, the pressure 1 atm of the gas equals the atmospheric pressure (1 atm), so, the force exerted by the gas (4.053 kN) and by the atmosphere (4.053 kN) on the side of the box balance each other. The net outward force is, then, zero. Is this right?
At 400 K, the pressure inside the box is 1.3 atm, which is bigger than the atmospheric pressure; the force exerted is bigger (5.4 kN) than the force exerted by the atmosphere (which continues 4.053 kN). The net outward force is, then, 5.4 - 4.053 = 1.347 kN. Is this right?
Thank you in advance.

Last edited: Oct 24, 2009
9. Oct 24, 2009

### Andrew Mason

Apart from your subtraction, yes. 5.4-4.05 = 1.35 kN

10. Oct 24, 2009

### pc2-brazil

OK, thank you.
I corrected the subtraction; I was in a hurry, so I mistyped it and didn't check (I did 5.4 - 4.53; I skipped the zero).
In this case, it seems that the question is ambiguous, because, for me, the net force is the sum of all forces acting in a body. In this case, I could simply say zero.

11. Oct 24, 2009

### Andrew Mason

The question would be pointless, if that was what it was asking. In the real world it is important to know what the net outward force is on the container - for example, in order to design a container with sufficient strength to sustain that force.

AM

12. Oct 25, 2009

### pc2-brazil

OK. It is very interesting.
This forum is very good for someone trying to self-teach. The book I have doesn't have the answer key to this question.