# Force exerted on inside of a box by a gas

## Homework Statement

A cubic metal box that has 27.8-cm (.278 m) long edges contains air at a pressure of 1.00 atm (~101,300 Pa) and a temperature of 303 K. The box is sealed so that the enclosed volume remains constant, and it is heated to a temperature of 416 K. Find the force due to the internal air pressure on each wall of the box

## Homework Equations

1. (P1*V1)/T1 = (P2*V2)/T2
2. P = F/A
3. Area of one side = length * width

## The Attempt at a Solution

I have no idea why I am not getting the correct answer here... From other sources I've found online, it looks like my methodology is correct.

Since the amount of gas is constant, I used the form of the ideal gas law given in equation 1. Since volume is constant, I canceled it from both sides of the equation, which leaves me with 101300 Pa/303 K = P2/ 416 K, which gives P2 = 139078 Pa. The area of one side is .278^2 = .077284 m^2. Solving for force in equation 2 and plugging in these numbers gives F = 139078 Pa * .07728 m^2 = 10747 N. However, this isn't the right answer.

Did I do something wrong here? This method seems to make sense, but doesn't yield the correct answer.

## Answers and Replies

never studied pressure before, but don't you have to consider the area of the entire box when calculating the force?
P = F / (area of entire box) ?

I did consider it, and maybe I should have mentioned that in my first post. Since the container in question is a cube, the total area of the inside is six times the area of one side (since a cube has 6 faces). That would give F = 139078 * 6*.077284 = 64487 N. (this isn't the correct answer either). Since I just want the force exerted on each wall (i.e. just one of the walls), I would have to divide that force by six. Dividing by six just gives me the same answer as I listed in the original post.

well, what is the right answer then? I get 64 507N, my only other guess is if you do temperatures in celcius, then it's 224 850 N
(edit: these need to be divided by 6 i forgot that part)

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The thing is, I don't know the right answer. Are you familiar with LON-CAPA at all? The homework problems are presented on this online service called LON-CAPA. You input your answer and it tells you whether you are right or wrong. You don't know what the right answer is until you enter it. We get ten tries at each problem, so I usually try a few things if I don't get the problem right away. Typically, if your answer is close (say, within 10, sometimes more, sometimes less depending on the problem) it will count the problem as correct.

Edit: I checked this using temperatures in celcius, still no dice.