What are some common misconceptions in thermodynamics?

For number 2... you need to use Q=mc(delta T) and also the ideal gas law PV = nRT to solve for the unknown variables and find the work done by the gas and the change in its internal energy.For number 3... you need to use the formula H=kA(change in T/L) to calculate the rate of energy transfer by conduction through the window. You only need to consider the air in the middle of the two panes of glass because it is the primary conductor of energy. The change in temperature will be the same for all layers since they are in thermal equilibrium.For number 4... you need to use the formula W=Pln(Vf/Vi) to calculate
  • #1
Kawrae
46
0
Sorry for some many questions here. I'm stuck on all of these problems and it's starting to really frustrate me. I'm not sure what more to do.

1. A sample of Helium behaves as an ideal gas as energy is added by heat at constant pressure from 273 K to 403 K. If the gas does 20.0 J of work, what is the mass of helium present?

>> At first I was going to use Q=mct, but I do not know c of helium and it's not in my physics book. Then I tried to use Q=mL which I think should work, but I don't know how to convert W to Q.

2. A gas is compressed from 10.00 L to 3.00 L at a constant pressure of 0.800 atm. In the process, 330 J of energy leaves the gas by heat. What is the work done by the gas? What is the change in its internal energy?

>> I know trial and error isn't the way to approach physics, but I'm lost. I tried using W=Pln(Vf/Vi) and got an answer of .963 J. I don't understand why this is wrong...

3. A thermal window with an area of 6.00m^2 is constructed of two layers of glass, each 4.00 mm thick and separated from each other by an air space of 4.00 mm. If the inside is at 20.0 C and the outside is at -34.0 C, what is the rate of energy transfer by conduction through the window?

>> I know I need to use the formula H=kA(change in T/L). But... do I have to do this using only the air in the middle of the two panes of glass? Or do I have to do the formula three times: one for the air, and two each piece of glass? And if so, is the change in temperature the same?

4. Gas in a container is at a pressure of 1.40 atm and a volume of 6.00 m^3. What is the work done by the gas if it expands at a constant pressure to twice its initial volume.

>> I think I need to use W=Pln(Vf/Vi). When I did this, though, I got .970... which isn't the answer according to webassign. I honestly don't know why this doesn't work. Does P have to be in a different unit?
 
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  • #2
Kawrae said:
Sorry for some many questions here. I'm stuck on all of these problems and it's starting to really frustrate me. I'm not sure what more to do.

1. A sample of Helium behaves as an ideal gas as energy is added by heat at constant pressure from 273 K to 403 K. If the gas does 20.0 J of work, what is the mass of helium present?

>> At first I was going to use Q=mct, but I do not know c of helium and it's not in my physics book. Then I tried to use Q=mL which I think should work, but I don't know how to convert W to Q.

2. A gas is compressed from 10.00 L to 3.00 L at a constant pressure of 0.800 atm. In the process, 330 J of energy leaves the gas by heat. What is the work done by the gas? What is the change in its internal energy?

>> I know trial and error isn't the way to approach physics, but I'm lost. I tried using W=Pln(Vf/Vi) and got an answer of .963 J. I don't understand why this is wrong...

3. A thermal window with an area of 6.00m^2 is constructed of two layers of glass, each 4.00 mm thick and separated from each other by an air space of 4.00 mm. If the inside is at 20.0 C and the outside is at -34.0 C, what is the rate of energy transfer by conduction through the window?

>> I know I need to use the formula H=kA(change in T/L). But... do I have to do this using only the air in the middle of the two panes of glass? Or do I have to do the formula three times: one for the air, and two each piece of glass? And if so, is the change in temperature the same?

4. Gas in a container is at a pressure of 1.40 atm and a volume of 6.00 m^3. What is the work done by the gas if it expands at a constant pressure to twice its initial volume.

>> I think I need to use W=Pln(Vf/Vi). When I did this, though, I got .970... which isn't the answer according to webassign. I honestly don't know why this doesn't work. Does P have to be in a different unit?

For 1... you know that P(deltaV) = work with constant pressure...

you also know that PV = nRT...

so PV1 - PV2 = nRT1 - nRT2 = P(deltaV) = nR(delta T) = work

you can figure out the number of moles... which you can use to determine the mass..
 
  • #3
1. The number c you looking for is 5.193 J/(grams*K)
Q=mc(delta T)
where delta T = 403 - 273

2. PV = nRT
If you change Volume, and Pressure stays the same - something's got to give. So T must rise.

3. http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html
This should help you understand difference between convection and conduction
 
Last edited by a moderator:
  • #4
cronxeh said:
1. The number c you looking for is 5.193 J/(grams*K)
Q=mc(delta T)
where delta T = 403 - 273

2. PV = nRT
If you change Volume, and Pressure stays the same - something's got to give. So T must rise.

3. http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html
This should help you understand difference between convection and conduction


For number 1... even if you use that equation... how are you going to solve it?
that gives you the energy change ... what good does that do? If a problem doesn't give you the value... i don't think it is needed to solve the problem, since enough information is available...
 
Last edited by a moderator:
  • #5
Kawrae said:
Sorry for some many questions here. I'm stuck on all of

2. A gas is compressed from 10.00 L to 3.00 L at a constant pressure of 0.800 atm. In the process, 330 J of energy leaves the gas by heat. What is the work done by the gas? What is the change in its internal energy?

>> I know trial and error isn't the way to approach physics, but I'm lost. I tried using W=Pln(Vf/Vi) and got an answer of .963 J. I don't understand why this is wrong...

For 2... P(deltaV) = work done on the system which is equal to the total energy...

q(heat) + Work = total energy... <--- solve for Work...
 
  • #6
cronxeh said:
2. PV = nRT
If you change Volume, and Pressure stays the same - something's got to give. So T must rise.

I understand why the temperature must rise... but how does that help me find the work done by the gas? :confused:
 
  • #7
If anyone helping me on this has AIM, can you IM me at ecksCavator? I'm so lost...
 
  • #8
Kawrae said:
Sorry for some many questions here. I'm stuck on all of these problems and it's starting to really frustrate me. I'm not sure what more to do.



3. A thermal window with an area of 6.00m^2 is constructed of two layers of glass, each 4.00 mm thick and separated from each other by an air space of 4.00 mm. If the inside is at 20.0 C and the outside is at -34.0 C, what is the rate of energy transfer by conduction through the window?

>> I know I need to use the formula H=kA(change in T/L). But... do I have to do this using only the air in the middle of the two panes of glass? Or do I have to do the formula three times: one for the air, and two each piece of glass? And if so, is the change in temperature the same?

4. Gas in a container is at a pressure of 1.40 atm and a volume of 6.00 m^3. What is the work done by the gas if it expands at a constant pressure to twice its initial volume.

>> I think I need to use W=Pln(Vf/Vi). When I did this, though, I got .970... which isn't the answer according to webassign. I honestly don't know why this doesn't work. Does P have to be in a different unit?

P should be in Pascals... for P(deltaV) to work... However though I'm not sure about W = Pln(Vf/Vi), but you can try it out...

I haven't done thermodynamics for energy transfer... so i can't help you on 3...
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the relationships and conversions between heat and other forms of energy.

2. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern energy and its transformations. They include the first law, which states that energy cannot be created or destroyed, only transferred or converted, and the second law, which states that the total entropy of a closed system always increases over time.

3. How is thermodynamics used in everyday life?

Thermodynamics has many practical applications in everyday life, such as in heating and cooling systems, engines, refrigerators, and power plants. It also plays a crucial role in understanding weather patterns, chemical reactions, and biological processes.

4. What is entropy?

Entropy is a measure of the disorder or randomness of a system. In thermodynamics, it is often referred to as the measure of the unavailable energy in a closed system.

5. How does thermodynamics relate to the concept of energy?

Thermodynamics and energy are closely related, as thermodynamics is the study of energy and its transformations. The laws of thermodynamics govern the flow and conversion of energy, and understanding thermodynamics is essential for understanding energy and its impact on our world.

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