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
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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
 
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Looks good to me. Why the doubtful question mark ?
 
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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 !
 
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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
 
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