Thermodynamic process, gas expansion

Click For Summary
SUMMARY

The discussion centers on a thermodynamic process involving 1 kg of air at an initial pressure of 106 Pa and a temperature of 398 K, which expands to five times its initial volume. The key relationship established is that the absorbed heat energy (dQ) is one-fourth of the work done by the gas (dW). The user rules out isochoric, isobaric, and adiabatic processes, concluding that the process is not isothermal either due to the relationship between dQ and dT. The challenge lies in calculating the final pressure and understanding the implications of the work-energy relationship in this context.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with the concepts of internal energy (U) and work (W) in thermodynamics
  • Knowledge of gas laws and properties of diatomic gases
  • Ability to manipulate equations involving heat transfer and work done
NEXT STEPS
  • Study the derivation of the first law of thermodynamics and its applications
  • Learn about the specific heat capacities of diatomic gases and their implications
  • Explore the concept of work done during gas expansion and its path dependency
  • Investigate the relationships between pressure, volume, and temperature in thermodynamic processes
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling gas expansion problems, as well as educators looking for insights into common misconceptions in thermodynamic processes.

usn7564
Messages
61
Reaction score
0

Homework Statement


1kg air with the pressure 10^6 Pa and temperature 398K expands until the volume is 5 times its' initial value. The expansion occurs so that in every moment the absorbed heat energy (dQ) is 1/4 of the work done by the gas. Find the final pressure. 1kmol of air has the mass 29kg, [tex]C_v=\frac{5}{2}R[/tex]

The attempt at a solution
Made numerous but it eventually stopped being about physics ended up being a desperate plug and chug. More interested in learning where I'm going wrong with my general line of thought.

I'm ruling out that the process is isochoric or isobaric, the former by definition and the latter because the book barely has any trick questions and I also looked at the answer.
And it's not adiabatic because dQ=/=0. That leaves isothermic but as [tex]dQ = C_v \nu dT[/tex] it can't be isothermic either.
And if it's none of those I simply don't know how to calculate the work, and as the work depends on path I can hardly chose an arbitrary one.

The only step forward I can take is [tex]dU=\frac{3}{4}dW[/tex], as dQ = 1/4 dW. I thought of somehow calculating change in internal energy as it doesn't depend on path, but it does depend on temperature and I don't have dT (nor do I know how to get it).

Any input would be appreciated, thanks.
 
Last edited:
Physics news on Phys.org
Air is diatomic. Can you use that to express U in terms of n and T (or better, express U in terms of P and V)? Also, I'm not sure about your sign for dU = (3/4)dW. Is dW the work done by the gas or the work done on the gas?
 

Similar threads

Closed cycle of an ideal gas
  • · Replies 3 ·
Replies
3
Views
2K
Thermodynamics Problem: Reversible and Irreversible Processes
  • · Replies 1 ·
Replies
1
Views
2K
Why is temperature constant after gas has expanded?
  • · Replies 33 ·
2
Replies
33
Views
3K
Is Entropy decreased for Free expansion of a Waals gas?
  • · Replies 6 ·
Replies
6
Views
4K
Thermodynamic Work vs Mechanical Work
  • · Replies 3 ·
Replies
3
Views
1K
Efficiency of Stirling Cycle - Is it the same as the Carnot Engine?
  • · Replies 14 ·
Replies
14
Views
2K
Undergrad Graphical difference between adiabatic and isothermal processes.
  • · Replies 1 ·
Replies
1
Views
4K
Find Isobaric Expansion & Pressure-Volume Coefficient for Solid
  • · Replies 4 ·
Replies
4
Views
2K
Internal Energy of an Ideal Gas
  • · Replies 4 ·
Replies
4
Views
3K
First law of thermodynamics & state functions
  • · Replies 16 ·
Replies
16
Views
2K