Calculating Work Done on a System with Changing Pressure and Volume

[sweetpete28]

n = 9.95 moles of ideal gas are slowly heated from initial pressure Po = 93398.6 Pa and initial volume Vo = .679 m^3 to final pressure Pf = 117508 Pa and final volume Vf = 1.073 m^3. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system:

If the plot is a straight line then it is an isobaric process so W = nRTb (1 - Va/Vb)

Using n = PV/RT, Tb=1524.172269K

W = (9.95)(8.314)(1524.172269)(1 - .679/1.073) = 46.3 kJ

Can someone please advise why/how this is wrong??

 

[sweetpete28]

Really trying here and really struggling...please, please help.

 

[Andrew Mason]

n = 9.95 moles of ideal gas are slowly heated from initial pressure Po = 93398.6 Pa and initial volume Vo = .679 m^3 to final pressure Pf = 117508 Pa and final volume Vf = 1.073 m^3. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system:

If the plot is a straight line then it is an isobaric process so W = nRTb (1 - Va/Vb)

It did not say horizontal line. Plot the two points on a PV diagram. They are not at the same pressure so how can the process be isobaric?

Is there a geometric aspect of the graph that represents the work done? What is it? Why?

AM

 

[sweetpete28]

The area under the curve but that is not correct either.

 

[Andrew Mason]

The area under the curve but that is not correct either.

What is the mathematical equation for the area under the graph of a straight line that is not parallel to the bottom axis?

AM

 

[sweetpete28]

Area of the rectangle + Area of the triangle.

 

[Andrew Mason]

Area of the rectangle + Area of the triangle.

So plot the two points, draw the line between them (straight) and work out the area. That is the amount of work done by the system. Caution: be careful to get the sign right. The question asks for the work done ON the system. How is this related to the work done BY the system?

AM

 

[RTW69]

It would appear that W_rev=W_{on gas}-W_{by gas}. We can calculate W_{by gas} but since the system is slowly heated we can't assume it is adiabatic. I don't see how we calculate W_rev and without we can't calculate W_{on gas}. What am I missing?

 

[Andrew Mason]

It would appear that W_rev=W_{on gas}-W_{by gas}.

This is not correct. The process is very slow so it will be close to a reversible process so W_rev ≈ W_actual ≈ ∫PdV where W is the work done by the gas in the reversible and actual situation. If internal pressure of the gas is infinitessimally higher than the external pressure on the gas for the duration of the process the reversible and actual work done would be equal.

We can calculate W_{by gas} but since the system is slowly heated we can't assume it is adiabatic. I don't see how we calculate W_rev and without we can't calculate W_{on gas}. What am I missing?

Calculate ∫PdV where P is the internal pressure of the gas which is determined by the co-ordinates of the PV diagram at all points between the beginning and end points that are given.

AM

 

[Chestermiller]

The temperature is controlled in such a way that the pressure versus volume plot is a straight line. You can use the ideal gas law to determine precisely how the temperature is being controlled. Lucky for you, the p vs v plot is a straight line.

Chet

 

[plazprestige]

The process you described is not isobaric: isobaric means the pressure is constant, but the pressure - as you said in the problem - changes.

The work associated with a system like this is equal to the negative area of the P-V graph. Negative work means that work is done by the system, and positive work means work is done on the system.

Sounds like you're finding the area of a trapezoid. Your trapezoid height will be the difference between the volumes (Vf - Vi), one base will be (Pi - 0) and the other base will be (Pf - 0).