# Homework Help: Ideal gas problem

1. Jul 26, 2004

### daisyi

Okay, I've been working on this for a while:

Three-tenths mole of an ideal gas at 400K is expanded isothermally from a pressure of 5x10^5 N/m^2 to 1.5x10^5N/m^2. If the gas is then heated to 500K at constant volume and then is compressed isothermally back to 5x10^5 N/m^2, and then isobarically back to the original condition, find the net work of the cycle.

I understand the whole concept of expanding and contracting, but combining the ways that these happen is very confusing to me.

first i found the volume for the first part using the formula V = nRT/P.
so for the first part, when its expanded isothermally i used the formula:
w = nRTln(V2/V1)
w= (3/10)*(8.314)*(400)*ln(6.65*10^-3/2*10^-3)
= 1198.68

same for the second isothermic expansion, except the temperature changed to 500, and therefrore the volume changed and:

w = -1503.58

then for the third part, i used the formula for isobaric gas:

w = P(V2 - V1)
= 245

Then for the net work, I added everything together, and got -364.8

What am I missing here??

Thanks!

2. Jul 26, 2004

### drag

If you have a set temprtature and pressure, the volume is
also determined for an equal amount of gas. So, if at the
end you get the original condition - all of the same parameters,
what kind of energy difference = work, are you looking for ?

3. Jul 26, 2004

### daisyi

it just says find the net work. I stated the problem exactly as it is written. Grr...this one is driving me nuts!

4. Jul 26, 2004

### Staff: Mentor

Looks good. (You are defining work as work done by the gas on the environment. So + work means that energy is leaving the gas. That's OK. Just be aware that some define work as work done on the gas.)

Take a second look at this one. I suspect you are using the wrong final volume.

5. Jul 26, 2004

### PureEnergy

daisyi - if you add up all your work terms again, you'll see it will not give you -365. The calculation for all the individual work terms are correct, except for some rounding errors. When I did the calcuations I get: 1201.18 - 1501.47 + 249.42 = -50.87 or -50 after rounding.

6. Jul 27, 2004

### Staff: Mentor

one more thing...

I never got to this step in my previous post:
It's an isobaric compression: the work will be negative.

7. Jul 27, 2004

### daisyi

I double checked and it still looks right to me. This is how I did it.

V = nRT/P

so, when the gas is heated to 500k with constant volume, the volume is:

V1 = (3/10)(8.314)(500)/1.5x10^5
= 8.314x10^-3

and then when the pressure is compressed back to 5x10^5:

V2= (3/10)(8.314)(500)/5*10^5
= 2.49x10^3

I then plugged that into the equation w = nRTln(v2/v1) to get the 1198.68

8. Jul 27, 2004

### Staff: Mentor

Right.

No. V1 = (3/10)(8.314)(400)/1.5x10^5 (You know the temp and pressure before it's heated.)

Right.

9. Jul 27, 2004

### PureEnergy

Doc's right, the third step should be negative so the net work is around -550.