# Homework Help: Easy thermo problem: gas expansion

1. Sep 3, 2007

### EricVT

1. The problem statement, all variables and given/known data

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

(a) Sketch a graph of pressure versus volume for this process.

(b) Calculate the work done on the gas during this process, assuming that there are no other types of work being done.

(c) Calculate the change in the helium's energy content during this process.

(d) Calculate the amount of heat added to or removed from the helium during this process.

2. Relevant equations

dQ = W + Q

W = - integral from intial volume to final volume of P(V)dV

3. The attempt at a solution

(a) The graph is just a linear curve with slope 1 from the point (1,1) to (3,3) on the P(atm) vs. V(L) axis.

(b) In SI units we have P(V)= 1.013x10^8 * V (from Pi/Vi and Pf/Vf) so the total work done on the gas is

W= - integral from .001 to .003 of (1.013x10^8)(V)(dv) and this I found to be -405.2 Joules

(c) Since I can't see that any heat was added to or removed from the system is the change in energy just -405.2 Joules?

(d) The heat added to or removed is just zero, isn't it? I don't understand how to know othewise. Unless the net change in energy is zero, then I don't see why heat would be added or removed.

Am I way off here?

2. Sep 4, 2007

### chaoseverlasting

You've done everything right as far as I can see.

3. Sep 4, 2007

### EricVT

Actually I did a little more work and the solution I ended up handing in was a bit more complicated.

Since Helium a single atom element, it only has translational kinetic energy contributions, so each Helium atom contributes (3/2)(k)(T) energy to the collective. The entire initial energy would be (N)(3/2)(k)(T)=(3/2)(P)(V) from the ideal gas law. We have both initial pressure and volume, so energy can be calculated here.

Then after the expansion takes place we have the same expression except we plug in our new values of P and V to get final energy. The difference is the change in energy of the system. The work done (calculated above) and the heat added have to equal this change in energy, so heat can be found as such.

Hopefully my second guess/second attempt is right, and it wasn't as easy as I originally thought :/

4. Sep 4, 2007

### learningphysics

You can also try using the specific heat capacity of helium, to calculate the heat entering the system.