How high will the piston rise if its temperature is increased?

  • Thread starter Thread starter kagnaroos
  • Start date Start date
  • Tags Tags
    Thermodynamics
Click For Summary

Homework Help Overview

The problem involves a closed cylinder with a massless piston and a gas, where the initial conditions include pressure, volume, and temperature. The question seeks to determine how high the piston will rise when the temperature is increased.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationships between pressure, volume, and temperature using the ideal gas law. There is uncertainty about the correctness of expressions for pressure and volume as functions of displacement. Some participants express doubt about the complexity of the resulting equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's expressions and suggesting corrections. There is a recognition of the complexity involved in deriving the relationships, and some guidance has been offered regarding the use of quadratic equations.

Contextual Notes

Participants note the absence of numerical data and express concern about the feasibility of solving the problem within a limited time frame, referencing its origin as an exam question.

kagnaroos
Messages
3
Reaction score
0

Homework Statement



[PLAIN]http://img101.imageshack.us/img101/7498/thermo.jpg

A cylinder is closed by a massless piston of cross sectional area A which is connected to a spring of constant k. With the spring relaxed, the cylinder is filled with a gas. The initial pressure, volume and temperature are Pi Vi and Ti. How high will the piston rise if its temperature is increased to a final value of Tf?

Homework Equations



The Attempt at a Solution



I tried to find Pf and Vf.

Pf = Pi + kx/A
Vf = Vi + Ax

I then wrote the universal gas law for Pf Tf and Vf but I doubt this is correct because the equation was crazy long and complicated.
 
Last edited by a moderator:
Physics news on Phys.org
In spite of being crazy long ad complicated, it might be correct. There were no numerical data?

ehild
 
Nope, only letters. Are the expressions I wrote for Pf and Vf correct?
 
kagnaroos said:
...

I then wrote the universal gas law for Pf Tf and Vf but I doubt this is correct because the equation was crazy long and complicated.
You have written out the expression for the pressure as a function of x and volume as a function of x. All you have to do is relate that, using the ideal gas law, to the change in temperature.

AM
 
kagnaroos said:
Nope, only letters. Are the expressions I wrote for Pf and Vf correct?

I think they are correct. But the expression for x will be really ugly. Just go ahead.

ehild
 
That gives a quadratic equation:

kx2 + (PiA + kVf/A)x + nR(Ti-Tf) = 0

I highly doubt this is the correct answer because this is an old exam question and impossible to do in ~10 minutes.
 
Well, it is not quite correct, there should be Vi instead of Vf in your term kVf/A. And you should replace nR by piVi/Ti.

Then use the formula for the solution of quadratic equations and exclude the negative root. It takes less time than 20 s.

ehild
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
6K