Me again (ideal gas/free expansion) soon

  • Thread starter don_anon25
  • Start date
  • Tags
    Expansion
In summary, the final temperature and pressure of the gas will be 45 degrees Celsius and 1 bar, respectively, and no entropy will be produced.
  • #1
don_anon25
36
0
Here's my problem:
An ideal gas is allowed to expand freely into an evacuated chamber.
The gas is at a pressure of 5 bar and a temperature of 45 degrees Celsius.
Find the final temperature and pressure of the gas.
Find the entropy produced.

W=0, Q=0, dU=0, and dT=0. So the final temp. is 45 degrees Celsius.
I need pressure at final state though...
dU=0 implies that u1=u2. But I don't know what the gas is, so I can't get values for these.

Am I missing something obvious?

Please help soon!
Thanks!
 
Physics news on Phys.org
  • #2
Since the gas is allowed to expand freely into an evacuated chamber, the final pressure of the gas will be equal to the atmospheric pressure. Therefore, the final pressure of the gas will be 1 bar. The entropy produced in this process can be calculated using the equation ∆S = ∆Q/T, where ∆Q is the heat transfer and T is the temperature of the system. Since no heat transfer occurs in this process, the entropy produced is 0.
 
  • #3


Hi there! It seems like you have the right idea so far. Since this is a free expansion, there is no work being done (W=0) and no heat being transferred (Q=0). This means that the change in internal energy (dU) is also 0.

You are correct that this implies that the initial internal energy (u1) is equal to the final internal energy (u2). However, since we don't know the specific gas being used, we can't determine the values for u1 and u2.

But, we can still find the final pressure and temperature using the ideal gas law: PV = nRT. Since the gas is expanding freely, the volume (V) increases while the number of moles (n) and the gas constant (R) remain constant. This means that the pressure (P) and temperature (T) must also change to maintain the equation.

To find the final pressure, we can use the initial pressure (5 bar) and volume (unknown) to solve for the final pressure. Similarly, we can use the initial temperature (45 degrees Celsius) and volume (unknown) to solve for the final temperature.

As for the entropy produced, we can use the equation ΔS = nRln(V2/V1) to calculate it. Again, since the number of moles and gas constant remain constant, we can use the initial and final volumes to find the change in entropy.

I hope this helps! Let me know if you have any other questions. Good luck!
 

1. What is an ideal gas?

An ideal gas is a theoretical gas that follows the ideal gas law, which states that the pressure, volume, and temperature of a gas are directly proportional to each other. This means that as one of these variables changes, the others will also change in a predictable manner.

2. How does free expansion occur in an ideal gas?

Free expansion in an ideal gas occurs when the gas is allowed to expand into a vacuum without any external work being done on it. This results in an increase in volume and decrease in pressure, as the gas molecules spread out to fill the available space.

3. What is the significance of free expansion in thermodynamics?

Free expansion is significant in thermodynamics because it demonstrates the concept of entropy, which is a measure of the disorder or randomness of a system. In free expansion, the gas molecules become more disordered as they spread out, leading to an increase in entropy.

4. Can an ideal gas ever truly experience free expansion?

No, an ideal gas can never truly experience free expansion because there will always be some external forces, such as gravity, acting on the gas molecules. However, in certain conditions, the effects of these external forces can be negligible and the gas can behave as if it is undergoing free expansion.

5. How does the temperature of an ideal gas change during free expansion?

The temperature of an ideal gas will remain constant during free expansion, as there is no external work being done on the gas to change its internal energy. This is known as an adiabatic process, where there is no transfer of heat.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
784
  • Introductory Physics Homework Help
Replies
8
Views
745
  • Introductory Physics Homework Help
Replies
33
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
491
  • Introductory Physics Homework Help
Replies
2
Views
540
  • Introductory Physics Homework Help
Replies
4
Views
951
  • Introductory Physics Homework Help
Replies
10
Views
916
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top