# [Thermo] One m^3 of an ideal gas expands in an isothermal

1. Sep 26, 2015

### leafjerky

1. The problem statement, all variables and given/known data
One m3 of an ideal gas expands in an isothermal process from 760 to 350 kPa. Determine the specific work done by the gas.

2. Relevant equations
ω=W/m
1W2=mRTln(P1/P2) = P1V1ln(P1/P2)

P1V1=P2V2

3. The attempt at a solution
P1V1ln(P1/P2) = (760)(1)ln(760/350) = 589.29kJ

V2=(760)(1)/(350) = 2.17m3

I tried doing PV=mRT using STP and my mass calculated was .31kg but it doesn't seem right. How do I find mass?

2. Sep 26, 2015

### davidbenari

Are you sure you're not given the molar mass of the substance? (or allowed to suppose its a certain element, like Nitrogen for example). Also, in case you didn't know in $PV=mRT$ , $m$ referes to moles and not mass... So its more usually written as $PV=nRT$

3. Sep 26, 2015

### leafjerky

Oh alright I'll remember that. I just have it written down differently. Unfortunately, that's all I'm given. Maybe there was a typo and she meant to just say work instead of specific work?

4. Sep 26, 2015

### davidbenari

Maybe she meant work per mole, which you can find.

5. Sep 26, 2015

### leafjerky

How?

6. Sep 26, 2015

### davidbenari

Well work along an isothermal path is $W_{1\to2} = -nRT \ln\frac{V_2}{V_1}$. You can see that $\frac{W{1\to2}}{n}=-RT \ln\frac{V_2}{V_1}$ would be the work done per mole.

You can do manipulations like the ones you did in your original post (such as P1V1=P2V2) to figure out more compact expressions that match the given data.

7. Sep 28, 2015

### leafjerky

She said it was supposed to be 1 m3/kg, so how does this change it? I know that gives us specific volume, not sure what to do with it.

8. Sep 28, 2015

### davidbenari

Well I would say that makes no sense. Specific work has units J/kg.

9. Sep 28, 2015

### SteamKing

Staff Emeritus
Mass density, ρ, is the reciprocal of specific volume, v , such that ρ = 1 / v.

https://en.wikipedia.org/wiki/Ideal_gas_law

10. Sep 28, 2015

### leafjerky

Okay so if I have 1 m3/kg of an ideal gas, the mass density of that gas is 1 kg/m3, what do I do with that and how does that get me to find my work? I've manipulated the equations as much as I know how and now I'm even more lost than I was when I posted this problem.

11. Sep 30, 2015

### leafjerky

Hey guys, this still is unanswered, I could really use some help. Please?

Given:
specific volume (v1) = 1 m3/kg - ideal gas
P1 = 760 kPa
P2 = 350 kPa

Find specific work.

Sorry to nag, I just really want to know how to do this.

12. Sep 30, 2015

### davidbenari

In my opinion, you need someway to know the molar mass of this substance. The ideal gas equations don't care about how much mass you've got, but care about the number of particles; which is why it is expressed as PV=nRT where n is the number of moles.

I feel you're teacher is hiding something of great importance here. (The molar mass)

13. Sep 30, 2015

### davidbenari

I'm not sure you know any calculus yet but here is the reasoning for what I've just said:

Work is $-pdV$

Pressure is $P=\frac{nRT}{V}=\frac{\frac{m}{M}RT}{V}$

So work is $dW=-\frac{m}{M}RT\frac{dV}{V}$

And specific work is $d\omega =-\frac{1}{M}RT\frac{dV}{V}$

And we conclude you need the molar mass of the substance....

Are you sure that your problem doesn't say "1 m3 of Nitrogen gas..." or something along those lines?