The ideal gas law for an adiabatic process

AI Thread Summary
The discussion centers on calculating the final temperature of air in a diesel engine undergoing adiabatic compression. The initial temperature is given as 293K, and after compressing the air by a factor of 15, the final temperature is calculated to be 866K using the adiabatic process equation. There is a correction made regarding the exponent used in the equation, confirming that the correct factor for pressure increase is 2.95. Participants emphasize the importance of using the right equations and maintaining confidence in their calculations. The ideal gas law is also referenced to relate pressure, volume, and temperature in the process.
garyd
Messages
26
Reaction score
0

Homework Statement



Hi can someone please have a look at this question and let me know if I am on the right track, thanks.

A diesel engine requires no spark plug; instead the air in the cylinder is compresses so highly the fuel ignites spontaneously on injection to the cylinder.

Q. If the air is initially at 293K and then is compressed adiabatically by a factor of 15, what final temperature is attained ?( Just prior to injection, also consider air as an ideal gas)
Variables:
T1=293K
V1= 1
V2=1/15
T2=?

Homework Equations



T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1)

The Attempt at a Solution


Above relationship applies as process is adiabatic?

293× (1^(1.4-1)/ (1/15^(1.4-1) ))=866K
 
Last edited:
Physics news on Phys.org
Looks good to me. Why the doubtful question mark ?
 
  • Like
Likes 1 person
BvU said:
Why the doubtful question mark ?

Confidence, I am learning to trust the equations.

Another part of the question is; by what factor has the pressure increased? Can I use P V^Ƴ=constant, with p1=1, multiplied by ratio of V1/V2, both to the power of Ƴ? ans=2.95
 
Re post 1: T1 *V1^(Ƴ-1) = T1* V1^(Ƴ-1) I read T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1) which looks good.

Re post 3: numerically I get something different, but then I use ##\gamma##, not ##\gamma -1## as the exponent... Big difference !
 
  • Like
Likes 1 person
Re post 3: numerically I get something different said:
Yes I was using wrong exponent, 44 is the correct factor, thanks.
 
garyd said:
Confidence, I am learning to trust the equations.

Another part of the question is; by what factor has the pressure increased? Can I use P V^Ƴ=constant, with p1=1, multiplied by ratio of V1/V2, both to the power of Ƴ? ans=2.95

Once you have the temperature worked out, just use the ideal gas law: PV=nRT

P1V1/T1 = P2V2/T2

P2/P1 = (V1/V2)(T2/T1)

AM
 
  • Like
Likes 1 person
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top