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

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The discussion revolves around determining how high a piston will rise when the temperature of the gas inside a cylinder is increased. The participants are working through the relationships between pressure, volume, and temperature using the ideal gas law, while also considering the effects of a spring connected to the piston. There is some confusion regarding the expressions for pressure and volume, with one participant expressing doubt about the complexity of the equations derived. Another participant suggests that the quadratic equation formed can be solved quickly by excluding the negative root. The conversation emphasizes the importance of correctly applying thermodynamic principles to arrive at a solution.
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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.
 
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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
 
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