# Homework Help: Reversible Adiabatic Process Question

1. Aug 23, 2010

### kalbuskj31

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

An Ideal Gas at 300K has a volume of 15L and a pressure of 15 atm. Calculate the change when the system goes under a reversible adiabatic expansion to a pressure of 10 atm. Gamma = 5/3. Cv = 1.5R. q = 0(definition of adiabatic processes).

2. Relevant equations

dw = P DV (lowercase means squiggly d and uppercase means straight d)
P = nRT/V

3. The attempt at a solution

The answer in the book says w = 5130 J, which is the opposite of what I got for the internal energy.

First, I calculated n: n = PV/RT = 15 atm * 15 L / .08206 (L*atm / K * mole) * 300K

n = 9.14 moles

Then, I calculated Vf: Vf = [(Pf * Vi ^ (5/3)) / (Pf)] ^ 3/5

Vf = 19.13 L

dW = P DV
dw = NRT

(dw = P DV so then work = NRT * Integral [DV/V]) ****This is wrong since T changes temperature between states*****

w = nRT ln(Vf/Vi)

Plugged in everything and the answer didn't turn out.

Last edited: Aug 24, 2010
2. Aug 23, 2010

### Gear300

Your method looks to be fine, so you might want to consider the units. Rather than L, convert to m^3, and instead of atm, convert to pascals.

3. Aug 24, 2010

### kalbuskj31

Thanks for taking a look at the problem Gear300. The problem I found with my logic is that P, V, and T all are changing from state 1 to state 2 so I have to use the Internal Energy formula reversing the temperature. (i.e. Ti - Tf).

4. Aug 25, 2010

### Andrew Mason

Not sure what the question is. Are you trying to find the work done? Find the change in temperature and use:

$$W = \Delta U = nC_v\Delta T$$

AM

5. Aug 25, 2010

### kalbuskj31

andrew,

That's what I ended up doing to get the answer.