Calculating final pressure and temperature of cylinder

AI Thread Summary
The discussion focuses on calculating the final pressure and temperature of a cylinder undergoing rapid adiabatic compression with a compression ratio of 15, starting from a pressure of 101.325 kPa and a temperature of 298 K. Participants confirm that the process is adiabatic and utilize the equation P2/P1 = 15 to find P2, resulting in 1519875 Pa. They discuss the relationship between temperature and pressure using the formula T2/T1 = (P2/P1)^((n-1)/n), leading to a calculated final temperature of approximately 381.171 K. The conversation emphasizes algebraic manipulation of equations to isolate pressure and temperature variables, ensuring clarity in the calculations. The final results for pressure and temperature are confirmed through collaborative problem-solving.
mh1985
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Homework Statement



Using a value of n of 1.1, calculate the final pressure and temperature when the cylinder is compressed rapidly with a compression ratio of 15.

Starting pressure = 101.325 kPa
Starting temp = 298 K

Homework Equations



(V2/V1) = 15

The Attempt at a Solution



Not sure about this part
 
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Rapid compression would mean that the compression is adiabatic. So you also know what other equation?

The compression ratio would be P2/P1=15 and you know what P1 is.

Using the above equation you found and the ideal gas equation, can you form an equation involving only P and T?
 
rock.freak667 said:
Rapid compression would mean that the compression is adiabatic. So you also know what other equation?

The compression ratio would be P2/P1=15 and you know what P1 is.

Using the above equation you found and the ideal gas equation, can you form an equation involving only P and T?

Thanks for the reply,

So I multiply both sides of P2/P1=15 by P1 to get P2,

101325 * 15 = P2 = 1519875 Pa

Not sure how to form the equation involving only P & T...

Something like T2/T1 = (P2/P1)^((n-1)/n)

EDIT:

so (P2/P1)^((n-1)/n) = 15^(1.1 - 1)/1.1 = 1.2791

1.2791 = T2/T1

T1*1.2791 = T2 = 381.171 K

Not sure if this is right?
 
Last edited:
mh1985 said:
Thanks for the reply,

So I multiply both sides of P2/P1=15 by P1 to get P2,

101325 * 15 = P2 = 1519875 Pa

Not sure how to form the equation involving only P & T...

Something like T2/T1 = (P2/P1)^((n-1)/n)

EDIT:

so (P2/P1)^((n-1)/n) = 15^(1.1 - 1)/1.1 = 1.2791

1.2791 = T2/T1

T1*1.2791 = T2 = 381.171 K

Not sure if this is right?

Well you'd have P1Vn1=P2V22 and P1V1/T1 = P2V2/T2


Take the last equation, make V1 the subject and substitute it into the first,the V2 should at least cancel out if I remember correctly.
 
rock.freak667 said:
Well you'd have P1Vn1=P2V22 and P1V1/T1 = P2V2/T2


Take the last equation, make V1 the subject and substitute it into the first,the V2 should at least cancel out if I remember correctly.

Hi thanks for the reply

If I make V1 the subject, I get P2V2T1 /P1T2, right so far? :S

But I don't see how to substitute it into the first equation?

P1 * (P2V2T1 /P1T2) = P2V2^2
 
mh1985 said:
Hi thanks for the reply

If I make V1 the subject, I get P2V2T1 /P1T2, right so far? :S

But I don't see how to substitute it into the first equation?

P1 * (P2V2T1 /P1T2) = P2V2^2

Sorry, I meant your equation to sub into should be

P1V1n=P2V2n

Just sub your first equation and the volume should cancel out.
 
rock.freak667 said:
Sorry, I meant your equation to sub into should be

P1V1n=P2V2n

Just sub your first equation and the volume should cancel out.


So I have:

P1* (P2T1/P1T2)n= V2 ?
 
mh1985 said:
So I have:

P1* (P2T1/P1T2)n= V2 ?

You have

V_1= \frac{P_2 V_2 T_1}{T_2 P_1}

and

P_1V_1^n = P_2V_2^n

so that

P_1 \left( \frac{P_2 V_2 T_1}{T_2 P_1} \right)^n = P_2V_2^n

You can see that V2n is common to both sides and it will cancel out leaving you with only pressure and temperature variables. Thus you can easily get T2 with some simple algebra.
 
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