How Much Work Is Done in Adiabatic Compression of Gas?

Click For Summary

Homework Help Overview

The discussion revolves around the adiabatic compression of a gas, specifically focusing on calculating the work done during this process using the first law of thermodynamics. The problem involves a gas initially at a volume of 10^-3 m^3 and a pressure of 1 atm, which is compressed to half its volume, resulting in a pressure increase by a factor of three. The internal energy is defined in terms of pressure and volume.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the first law of thermodynamics, noting that the process is adiabatic and thus dQ = 0. There are attempts to differentiate the internal energy equation to find temperature, but some express uncertainty about the approach. Others question the rounding of results when calculating work done.

Discussion Status

Some participants are confirming their calculations and discussing minor discrepancies in the results, such as differences in rounding. There is acknowledgment of the correct application of principles, but no consensus on the exact numerical answer has been reached.

Contextual Notes

Participants are operating under the constraints of the problem statement, focusing on the definitions provided for internal energy and the conditions of the adiabatic process. The discussion reflects varying interpretations of the results based on rounding and calculation methods.

v_pino
Messages
156
Reaction score
0

Homework Statement


When a volume 10^-3 m^3 of a certain gas at a pressure of 1 atm undergoes a reversible adiabatic compression to half its volume, its pressure rises by a factor of three. The internal energy of the gas is given by E=3PV, where P is the pressure and V is the volume. By making use of the first law of thermodynamics, or otherwise, calculate how much work is done on the gas to perform the compression.

Answer: 152 J

Homework Equations


(1) I tried writing E=E(V,T) or E=E(P,T).

(2) First law of thermodynamics: dE=dQ+dW

The Attempt at a Solution


dQ = 0 for reversible process.
I differentiated (1) but had no way of getting T.

Am I on the right track?
 
Physics news on Phys.org
v_pino said:

Homework Statement


When a volume 10^-3 m^3 of a certain gas at a pressure of 1 atm undergoes a reversible adiabatic compression to half its volume, its pressure rises by a factor of three. The internal energy of the gas is given by E=3PV, where P is the pressure and V is the volume. By making use of the first law of thermodynamics, or otherwise, calculate how much work is done on the gas to perform the compression.

Answer: 152 J

Homework Equations


(1) I tried writing E=E(V,T) or E=E(P,T).

(2) First law of thermodynamics: dE=dQ+dW

The Attempt at a Solution


dQ = 0 for reversible process.
I differentiated (1) but had no way of getting T.

Am I on the right track?
Yes. You know that there was no heat input into the gas, because the process is adiabatic. Therefore the change in internal energy of the gas is entirely due to work done on the gas:

[tex]dE=dW[/tex]

Applying this statement, along with the equation given in the problem should be enough to get you to the answer.
 
I'm getting 150J instead of 152J as given in the answer. I substituted V1 and P1 into equation of E to get E1 and substituted V2 and P2 into equation of E to get E2.

E1 = 300 and E2 = 450.

Subtracting E1 from E2 gives 150J.
 
v_pino said:
I'm getting 150J instead of 152J as given in the answer. I substituted V1 and P1 into equation of E to get E1 and substituted V2 and P2 into equation of E to get E2.

E1 = 300 and E2 = 450.

Subtracting E1 from E2 gives 150J.

That's fine. You're doing the problem correctly. You're just rounding. Thus, the answer you get is rounded. I compute, using the same method:

E1=303.98J E2=455.96J Thus, W~152J
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K