Thermodoynamics

[moonlit]

I have 3 probs I'm working on, not really sure how to go about these...

1) A gas, while expanding under isobaric conditions, does 611 J of work. The pressure of the gas is 1.70 x 10^5 Pa, and its initial volume is 1.30 x 10^-3 m^3. What is the final volume of the gas?

I'm assuming I would use the equation: W=P delta V=P(Vfinal-Vinitial) but since I don't know the final volume how do I go about solving this problem?

2) Suppose 5.50 moles of a monatomic ideal gas expands adiabatically and its temperature decreases from 425 to 218 K. Determine (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy.

I thought that the correct way to go about this was to use W=3/2 nR(Tinitial-Tfinal) and I ended up with an answer of 7731.45 J for the first part of the problem but the program says it's wrong. How do I solve parts A and B?

3) One-half a mole of a monatomic ideal gas expands adiabatically and does 909 J of work. By how much does its temperature change in Kelvin? Use the proper sign to specify whether the change is an increase or a decrease.

For this prob I thought I could use the same equation as in #2 and I ended up with an answer of 24.34 but this sounds wrong. What should I do here?

[KLscilevothma]

Originally posted by moonlit
[B]I have 3 probs I'm working on, not really sure how to go about these...

1) A gas, while expanding under isobaric conditions, does 611 J of work. The pressure of the gas is 1.70 x 10^5 Pa, and its initial volume is 1.30 x 10^-3 m^3. What is the final volume of the gas?

I'm assuming I would use the equation: W=P delta V=P(Vfinal-Vinitial) but since I don't know the final volume how do I go about solving this problem?
The question asks you to find the final volume. The equation you mentioned is the one you need, as you've already got W, P and Vinitial.

2) Suppose 5.50 moles of a monatomic ideal gas expands adiabatically and its temperature decreases from 425 to 218 K. Determine (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy.

I thought that the correct way to go about this was to use W=3/2 nR(Tinitial-Tfinal) and I ended up with an answer of 7731.45 J for the first part of the problem but the program says it's wrong. How do I solve parts A and B?
I used W = 3/2 nR (Tinitial-Tfinal) and got 14198 J. I guess you might have used a wrong value of R.
The value of R is 3.814 Jmol-1K-1

Same for Q3, please check the numerical values you put into the equation.

[moonlit]

Hmmm, well I figured out the last problem but the first two I'm still having problems with. According to my formula sheet R=8.31 J/mol K NOT 3.814 so now I'm really confused. I'm also not sure how to figure out part b of question #2. Help me please!

[moonlit]

Ok, nevermind. I figured out problem #2. Thanks for the help! Still having trouble with the first one though...any pointers??

[Doc Al]

Originally posted by moonlit
Still having trouble with the first one though...any pointers??
What's the problem? As KLscilevothma has pointed out: You know the formula (W=PΔV) and you know all the variables except the one you are asked to find. (Of the three problems, this one is easiest.)

[moonlit]

Ok, I'm working on this problem:

1) A gas, while expanding under isobaric conditions, does 611 J of work. The pressure of the gas is 1.70 x 10^5 Pa, and its initial volume is 1.30 x 10^-3 m^3. What is the final volume of the gas?


Here's what I've done:
W=P delta V=P(Vfinal-Vinital)
611=1.7x10^5(Vfinal-1.30x10^-3)

But now I'm not sure how to go about solving it (can't remember the correct math steps). Can you help me?

[Doc Al]

Originally posted by moonlit
Here's what I've done:
W=P delta V=P(Vfinal-Vinital)
611=1.7x10^5(Vfinal-1.30x10^-3)

But now I'm not sure how to go about solving it (can't remember the correct math steps). Can you help me?

Schematically, it looks like this:

W = P(V2 - V1), so...

Divide by P:

W/P = V2 - V1

Now add V1 to both sides:

V2 = W/P + V1

Got it?

[moonlit]

Hmmmmm...I got an answer of 2.765915385 but the program is saying it's wrong.

Here's the work I did:

611/1.7x10^5=Vfinal-.0013
.003594/.0013=Vfinal-.0013
2.7646=Vfinal-.0013
2.7646+.0013=Vfinal-.0013+.0013
=2.7659 m^3

[Doc Al]

Originally posted by moonlit
Here's the work I did:

611/1.7x10^5=Vfinal-.0013
.003594/.0013=Vfinal-.0013
...

Why did you divide by .0013 on the left side of this equation?

[moonlit]

Shoot, nevermind...I figured it out. Thanks for the help!