Homework Help: Ideal Gas Exercise

1. Aug 12, 2008

Perrin

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

A cylinder with a frictionless piston is placed horizontally in an atmosphere pressure 1 * 105 N/m2. A gas in the cylinder is initially at a temperature of 300K with a volume of 6.0 * 10-3 m3. Then the gas is heated slowly to 400K. How much work is done by the gas in the process?

2. Relevant equations

Ideal Gas equations:

p*V/T = constant

p*V = NKT = nRT

Ek (average) = (3/2) * kT

Ek (total) = (3/2) * NkT = (3/2)pV

3. The attempt at a solution

At first I calculated the initial energy as:
E1 = (3/2)pV = (3/2)*(1 * 105) * (6.0 * 10-3) = 900J

Then, assuming the pressure is constant, I said:

pV1/T1 = pV2/T2

V2 = V1*T2 / T1

V2 = 6.0 * 10-3 * 400 / 300 = 8*10-3

Thus, the second energy:

E2 = (3/2)*105*8*10-3 = 1200J

And the work:
W = E2 - E1 = 1200 - 900 = 300J.

Now, in the answer to the question, it says the work is not 300J, but 200J.
Can someone enlighten me about my mistake?

Thanks.

2. Aug 12, 2008

Ygggdrasil

You calculated the change in internal energy of the gas (ΔE). ΔE = w + q, so you are ignoring q, the amount of heat transferred during the process. You should use this equation for work instead:

$$w = -\int_{v_1}^{v_2}{P_{ext}dV}$$

3. Aug 12, 2008

Perrin

So, I could also calculate:

w = ΔE - q ?

And could you please explain to me how to use the equation you wrote? What's Pext? How do you integrate it? Sorry, I'm not very good with integrals...

4. Aug 12, 2008

Ygggdrasil

Pext is the external pressure. If the external pressure stays constant during your process, P does not vary with V and you can pull it out of the integral to get the simpler relation w = -PextΔV. If P varies as the volume changes, you have to do the integration by finding an expression for P in terms of V.

5. Aug 12, 2008

Perrin

Thanks, I finally understand :D