Calculating final pressure and temperature of cylinder

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Discussion Overview

The discussion revolves around calculating the final pressure and temperature of a gas in a cylinder undergoing rapid adiabatic compression. Participants explore the relevant equations and relationships between pressure, volume, and temperature, particularly focusing on the implications of a compression ratio of 15 and an adiabatic process characterized by a specific heat ratio (n) of 1.1.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states the initial conditions: starting pressure of 101.325 kPa and starting temperature of 298 K, with a compression ratio of 15.
  • Another participant notes that rapid compression implies an adiabatic process and suggests using the relationship P2/P1 = 15 to find P2.
  • A participant calculates P2 as 1519875 Pa by multiplying the starting pressure by the compression ratio.
  • There is a proposal to use the equation T2/T1 = (P2/P1)^((n-1)/n) to find the final temperature, leading to a calculated T2 of approximately 381.171 K.
  • Further discussion includes manipulating equations involving volume and pressure, with participants attempting to isolate variables and substitute them into different equations.
  • One participant expresses uncertainty about the substitution process and seeks clarification on how to proceed with the algebraic manipulation.
  • Another participant provides guidance on how to cancel out volume terms in the equations to simplify the problem.

Areas of Agreement / Disagreement

Participants generally agree on the use of adiabatic relationships and the equations involved, but there is uncertainty regarding the algebraic steps and the correctness of the calculated values. The discussion remains unresolved as participants continue to seek clarification and refine their approaches.

Contextual Notes

Participants express uncertainty about specific algebraic manipulations and the implications of their calculations. There are unresolved steps in the derivation of relationships between pressure, volume, and temperature.

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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