Work and Volume in Adiabatic Expansion

Homework Statement

A sample of 1.60 mol H2 (Cv = 20.5 J K-1 mol-1) at 21°C and 1.50 atm undergoes a reversible adiabatic compression until the final pressure is 4.50 atm. Calculate the final volume of the gas sample and the work associated with this process. Assume that the gas behaves ideally.

Homework Equations

pv=nRT
p1v1=p2v2
ΔE=-p_eternalv=nC_vΔT

The Attempt at a Solution

pv=nRT → (1.5 atm)(V₀=(1.6 mol)(0.08206 atm mol K)(294 K)
V₀=25.734016 L (tried this initially as I was given Cv, but that was wrong)
P1V1=P2V2→(1.5 atm)(25.734016 L)=(4.5 atm)(V) →8.57800533 L (which was wrong)

ΔE=-(4.5)(17.1501067)= -77.202 J (nope)
ΔE=(1.6 mol)(20.5 K mol)(294 K)= 9643.2 J (Nope)

I'm not really sure where I went wrong with this, any input would be really helpful

 

[Chestermiller]

Is the process isothermal, or is it adiabatic?

 

Is the process isothermal, or is it adiabatic?

Adiabatic

 

[Chestermiller]

Adiabatic

For an adiabatic reversible process, is this equation correct: p1v1=p2v2 ?

 

For an adiabatic reversible process, is this equation correct: p1v1=p2v2 ?

That's what I wrote in my notes at least. I suppose not though?

 

[Chestermiller]

That's what I wrote in my notes at least. I suppose not though?

You suppose correctly. Do you know the correct equation? It can be derived by applying the 1st law of thermodynamics to this adiabatic reversible compression. What would the 1st law give you for this?

 

You suppose correctly. Do you know the correct equation? It can be derived by applying the 1st law of thermodynamics to this adiabatic reversible compression. What would the 1st law give you for this?

The first law of thermodynamics would be conservation of energy, which would be Internal Energy=Q+W. Since it's an adiabatic system, Energy would just be equal to work. E=p deltaV and nCv delta T... Right?

 

[Chestermiller]

The first law of thermodynamics would be conservation of energy, which would be Internal Energy=Q+W. Since it's an adiabatic system, Energy would just be equal to work. E=p deltaV and nCv delta T... Right?

Yes! Except that, because the process is reversible, those deltas should be d’s. That is, differentials. And there should be a minus sign. Now substitute p=nRT/V into the equation. What does that give you?

 

Do you mean the volume value I calculated before or the actual equation itself?
E= -(4.5 atm) (d/dv 25.734016 L) =...
E= nRT/v x d/dv V -> E= nRT d/dv
(I'm sorry I lost momentum here I'm unsure what to do.)

 

[Chestermiller]

What I'm saying is $$nC_vdT=-\frac{nRT}{V}dV$$ Does this equation make sense to you?

 

What I'm saying is $$nC_vdT=-\frac{nRT}{V}dV$$ Does this equation make sense to you?

I thought I did but based on what I came up with it seems I've shamed my calculus teacher

∫nC_vdt=-∫nRT/v dV → nC_vT₁-nC_vT₂=-(nRT)ln(V₁)-(nRTln(V₂)
So if I plug what I know
(1.6 mol)(20.5 J K-1 mol-1)(294K) - (1.6 mol)(20.5 J K mol)(T₂)=-[(1.6)(8.315)(294)ln(25.734016 L)-(1.6)(8.315)(294)(lnV₂)]

-3060.220545=(1.6 mol)(20.5 J K mol)(T) - (319.296)(ln (25.734016 L/ V₂)

So I'm assuming this is super wrong?

 

[Chestermiller]

This is not integrated correctly. And, please stop substituting numbers. Do it all algebraically.