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

[leafjerky]

Homework Statement

One m³ of an ideal gas expands in an isothermal process from 760 to 350 kPa. Determine the specific work done by the gas.

Homework Equations

ω=W/m
₁W₂=mRTln(P₁/P₂) = P₁V₁ln(P₁/P₂)

P₁V₁=P₂V₂

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?

 

[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 $PV=nRT$

 

[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?

 

[davidbenari]

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

 

[leafjerky]

How?

 

[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.

 

[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.

 

[davidbenari]

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

 

[SteamKing]

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.

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

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

 

[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.

 

[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.

 

[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)

 

[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?