Problem using Gay-Lussac's law and forces

In summary, to find the net force on each side of a cubic box filled with air at different temperatures, we must use the ideal gas law in conjunction with the equation PV = nRT and ensure accuracy in conversions and units.
  • #1
lm93
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Homework Statement


A cubic box of volume 4.0 multiplied by 10-2 m3 is filled with air at atmospheric pressure at 20°C. The box is closed and heated to 160°C. What is the net force on each side of the box?

Homework Equations


P1/T1=P2/T2
P=F/A

The Attempt at a Solution


So we can assume that volume is constant because the box is closed. P1=1 atm P2=unknown
T1=20C T2=160C. Using the above equation I got: 1/20=P2/160--> 1/20*160=P2-->P2=8 atm
Then to find the force I thought I should use F=PA. I converted 8 atm into N/m2 and got 810 600N/m2. For the area of one side of the box, I thought that one length of the cubic box was the cube root of 0.04 squared which I found to be 0.11696m2. So then, I put those numbers into the equation and got F=PA--> F=0.11696*810600--> F=94808.35N
But this was wrong

I have tried this problem many different times and keep getting it wrong. I was wondering if it matters the units you use for the first equation P/T=P/T. I thought as long as you kept them constant it didn't...but then when I tried it with different units eg. Kelvins instead of degrees Celsius I got a different answer.

Could someone please help? I would really appreciate it!
Thank you
 
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  • #2
for your post and for showing your attempt at solving the problem. It seems like you have the right idea, but there are a few things that could be causing your incorrect answer.

First, the equation P1/T1 = P2/T2 is known as the ideal gas law, and it is typically used to relate pressure and temperature for a gas in a closed system. In this case, the box is closed, but it is not a closed system since the air can expand and change volume. So, we cannot use this equation to find the pressure of the air inside the box.

Instead, we can use the ideal gas law in conjunction with the equation PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin. Since we know the volume and temperature of the gas before and after heating, we can use this equation to find the final pressure of the gas inside the box.

Second, when converting from atm to N/m2, you should get 810,600 N/m2, not 810,600,000 N/m2. This could be a simple calculation error, but it is important to double check your conversions and units to ensure accuracy.

Lastly, the area of one side of the box should be the surface area, not just one length of the cubic box. Since the box is a cube, all sides have the same area, which can be found by multiplying the length of one side by itself. So, the area of one side should be (0.04 m)^2 = 0.0016 m^2.

With these corrections, your final answer should be around 15,000 N for the net force on each side of the box. I hope this helps and good luck with your studies!
 
  • #3
for sharing your attempt at solving this problem. It seems like you have a good understanding of the equations involved, but there may be a few small errors in your calculations.

First, when converting from Celsius to Kelvin, you need to add 273 to your temperature values. So for T1 = 20°C, the corresponding temperature in Kelvin would be 293K, not 20K.

Second, when converting from atm to N/m2, you need to multiply by 101325, not 810600. So your pressure in N/m2 would be 8*101325 = 810600 N/m2.

Third, when calculating the area of one side of the box, you need to use the surface area formula for a cube, which is 6a^2, not the cube root of 0.04 squared. So the area would be 6*(0.04)^2 = 0.0096 m2.

With these corrections, your final answer should be F = 810600 N/m2 * 0.0096 m2 = 7785.6 N.

In general, it is important to pay attention to units and make sure they are consistent throughout your calculations. This can often be a source of errors in problem solving. Additionally, it is always a good idea to double check your calculations and make sure they make sense in the context of the problem. If you are still struggling with this problem, it may be helpful to review the concepts of Gay-Lussac's law and forces, and practice similar problems to build your understanding and confidence. Good luck!
 

1. What is Gay-Lussac's law and how does it relate to forces?

Gay-Lussac's law is a gas law that states the pressure of a gas is directly proportional to its temperature, assuming the volume and number of particles remain constant. This means that as the temperature of a gas increases, so does its pressure. In terms of forces, this law shows that an increase in temperature can lead to an increase in the force exerted by a gas.

2. How can Gay-Lussac's law be used to solve problems?

Gay-Lussac's law can be used to solve problems involving the relationships between temperature, pressure, and volume of a gas. By rearranging the formula for this law, we can solve for any of these variables given the other two. This can be useful in various scientific and engineering applications.

3. What are some real-life examples of Gay-Lussac's law and forces?

One example of Gay-Lussac's law and forces can be seen in the operation of a gas-powered engine. As the temperature in the cylinder increases due to combustion, the pressure of the gas also increases, which in turn creates the necessary force to move the pistons and power the engine.

4. Are there any limitations or exceptions to Gay-Lussac's law?

Yes, Gay-Lussac's law is only applicable to ideal gases, which follow certain assumptions such as having no intermolecular forces and occupying zero volume. In real gases, there may be deviations from this law at high pressures or low temperatures.

5. How does Gay-Lussac's law differ from other gas laws?

Gay-Lussac's law specifically focuses on the relationship between pressure and temperature, while other gas laws such as Boyle's law and Charles's law involve the relationships between pressure and volume, and volume and temperature, respectively. Additionally, Gay-Lussac's law only applies to a constant volume, while the other laws involve changes in volume.

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