Thermodynamics He expanding volume directly proportional to pressure

[relativespeak]

Homework Statement

Imagine some helium in a cylinder with an initial volume of 1 litre and an initial pressure of 1 atm. Somehow,
the helium is made to expand to a final volume of 3 litres, in such a way that its pressure rises in direct
proportion to its volume.

(b) Calculate the work done on the gas during this process, assuming that the process occurs quasi-statically,
and no other types of work are being done.
(c) Calculate the change in the internal energy of the helium during the process.
(d) Calculate the amount of heat added to or removed from the helium during this process,
(e) What would you have to do to cause the pressure to rise as the helium expands?
(f ) Describe a simple experimental setup in which this form of the expansion would occur, i.e., where the
pressure would rise proportional to the volume of the gas.

Homework Equations

W = -∫PdV
U = Q+W = (f/2)NkT

PV = NkT

The Attempt at a Solution

For the first part I used W = -∫PdV and P=mV where m is some proportionality constant. Then W=-∫mVdV= -(m/2)V^2 which equals -m*4E-6. I'm not sure if this work is correct.

Then for part c, U=(3/2)NkT=(3/2)PV=-(3/2)W (because helium has 3 degrees of freedom) and the heat is given by:
-(3/2)W=Q+W, Q=-(5/2)W. I'm just not sure if this is correct

 

[TSny]

Your work for (a) looks good. You should be able to determine the numerical value of the constant of proportionality m and thus determine a numerical value for the work. You should specify units in your answers.

For part (c), note that you are asked for the change in U. Also, it is not correct to equate PV with -W.

 

[relativespeak]

How can I find a numerical value for m without knowing the change in P?

The change in U would then be U=ΔPΔV=mΔVΔV=m(ΔV)^2?

 

[TSny]

What does m have to be so that P = 1 atm when V = 1 litre?

 

[TSny]

How can I find a numerical value for m without knowing the change in P?

The change in U would then be U=ΔPΔV=mΔVΔV=m(ΔV)^2?

No. ΔU means U_f - U_i = (3/2)(PV)_f - (3/2)(PV)_i.

 

[Radhakrishnam]

The process is easily visualized on a PV diagram. The path followed by the system is shown by a straight line with a positive slope (joining the points (3,3) and (1,1). Taking this line as the hypotenuse complete the right angled triangle The third vertex will be the point (1,3).

The area under the hypotenuse gives the heat supplied to the system. The area of the triangle gives the increase internal energy. the difference between the above two areas gives (the area under the straight line parallel to the volume axis) gives the work done by the system. You will get all areas ( and the corresponding quantities) in terms of liter atm.

(e) heat the gas
(f) keep the cylinder horizontally so that the piston moves horizontally. As it moves the piston rotates a drum over which a thread, to one end of which a mass is attached, is wound. As the drum rotates the weight is lifted, which gives a measure of the work.