So, the force on each wall of the box is 138 N.

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Homework Help Overview

The problem involves a cubic metal box containing air, which is heated, leading to a change in pressure and the calculation of the force exerted on the walls of the box. The subject area includes thermodynamics and fluid mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the ideal gas law and the calculation of pressure changes due to temperature increase. Questions arise regarding the need to account for external pressure and how to calculate the net force on the walls based on pressure differences.

Discussion Status

Some participants have provided guidance on the importance of considering external pressure in the calculations. There is an ongoing exploration of unit consistency and the need to recalculate values to ensure accuracy.

Contextual Notes

Participants note the necessity of using SI units for volume and pressure to derive the force in Newtons. There is mention of potential errors in numerical computations and the importance of correctly expressing atmospheric pressure in the appropriate units.

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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

**that isn't the correct answer
 
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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

[tex]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}[/tex]

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.
 

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