Work done as air expands in piston with spring

[Giuseppe]

There's a problem in my thermo class that I think I have the right answer for, but the value I get doesn't seem to make sense. The problem reads:

Warm air is contained in a piston-cyliner assembly oriented horizontally. The air cools slowly from an intial volume of .003 m^3 to a final volume of .002m^3. During the process, a spring exerts a force that varies linearly from an intial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is .018m^2. Friction between the piston and the cylinder wall can be neglected. For the air, determine the intial and final pressures, in kPa, and the work in kJ.

So I split the system up into two states.

State 1: V1 = .003 m^3 Fspring = 900N
State 2: v2 = .002 m^3 Fspring = 0N

I found the pressures for each state using a force balance equation and the relationship between force and pressure.

I came up with:

Fair = Fspring + Farm
(Pair)(Apiston) = Fspring + (Pair)(Apiston)

Using the values for State 1 and 2 I came up with 150 kPa for State 1 and 100 kPa for state 2.

Here's where I run into problems. Calculating the work.

First off I make the assumption that the magnitude of the work done as the spring moves is equal to the magnitude of the work the air does during this change.

Therefore, I used 1/2kx^2 for work done. I found the distance the spring moved using (V2-V1)/Apiston = delta X

Using this x value I went to Fspring = 900 = kx and found k.

From here I used evaluated 1/2kx^2 and I came up with an answer of 25Joules.

My question is, is the answer I getting make sense physically, because it seems like a really small answer, also are the assumptions I am making for work done correct, or is there another way I should go about this.

 

[siddharth]

There's a problem in my thermo class that I think I have the right answer for, but the value I get doesn't seem to make sense. The problem reads:

Warm air is contained in a piston-cyliner assembly oriented horizontally. The air cools slowly from an intial volume of .003 m^3 to a final volume of .002m^3. During the process, a spring exerts a force that varies linearly from an intial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is .018m^2. Friction between the piston and the cylinder wall can be neglected. For the air, determine the intial and final pressures, in kPa, and the work in kJ.

So I split the system up into two states.

State 1: V1 = .003 m^3 Fspring = 900N
State 2: v2 = .002 m^3 Fspring = 0N

I found the pressures for each state using a force balance equation and the relationship between force and pressure.

I came up with:

Fair = Fspring + Farm
(Pair)(Apiston) = Fspring + (Pair)(Apiston)

Using the values for State 1 and 2 I came up with 150 kPa for State 1 and 100 kPa for state 2.

That's correct.

Here's where I run into problems. Calculating the work.

First off I make the assumption that the magnitude of the work done as the spring moves is equal to the magnitude of the work the air does during this change.

Therefore, I used 1/2kx^2 for work done. I found the distance the spring moved using (V2-V1)/Apiston = delta X

Using this x value I went to Fspring = 900 = kx and found k.

From here I used evaluated 1/2kx^2 and I came up with an answer of 25Joules.

My question is, is the answer I getting make sense physically, because it seems like a really small answer, also are the assumptions I am making for work done correct, or is there another way I should go about this.

Now, while calculating the work done, you have to be careful. First of all, what work is the question asking for? I'll assume that the air in the cylinder is the system, and the work done on the system (by the surroundings, ie the atmosphere and the spring) in this process is what is required.

The only work done on the system is expansion work, which is \int P dv. So, let us say that the spring is compressed by a distance 'x'. Can you calculate the pressure in the cylinder as a function of 'x'? Note that this process has to be reversible, which means that the net force on the piston is always 0.

Once you do this, you will be able to calculate the work done, as 'x' goes from the initial value, to 0. Also, watch your signs when you calculate the final answer.