# Work and Volume in Adiabatic Expansion

• ScreamingIntoTheVoid
In summary, the conversation revolved around solving a problem involving the compression of a gas sample with given values for pressure, volume, and temperature. After attempting to apply various equations, it was determined that the correct approach was to use the first law of thermodynamics and substitute the ideal gas law into the equation. The final answer was not achieved due to incorrect integration and substituting numbers instead of solving algebraically.
ScreamingIntoTheVoid

## 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=-peternalv=nCvΔT

## 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

Is the process isothermal, or is it adiabatic?

Chestermiller said:
Is the process isothermal, or is it adiabatic?

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

Chestermiller said:
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?

ScreamingIntoTheVoid said:
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?

Chestermiller said:
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?

ScreamingIntoTheVoid said:
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.)

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

Chestermiller said:
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

∫nCvdt=-∫nRT/v dV → nCvT1-nCvT2=-(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?

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

## 1. What is adiabatic expansion?

Adiabatic expansion is a process in thermodynamics where the volume of a gas increases without any heat entering or leaving the system. This means that the temperature of the gas decreases, as the internal energy is used to do work on the surrounding environment.

## 2. How is work related to adiabatic expansion?

In adiabatic expansion, work is done by the gas as it expands against the external pressure. This work is equal to the negative of the change in internal energy of the gas, as there is no heat exchange involved.

## 3. What is the formula for calculating work in adiabatic expansion?

The formula for work in adiabatic expansion is W = -PΔV, where W is work, P is pressure, and ΔV is the change in volume of the gas.

## 4. How does the volume of a gas change during adiabatic expansion?

The volume of a gas increases during adiabatic expansion, as the gas does work on the surrounding environment and releases internal energy. This results in a decrease in temperature, as the gas's molecules have less kinetic energy.

## 5. What is the relationship between work and temperature in adiabatic expansion?

In adiabatic expansion, the temperature of the gas decreases as work is done on the surrounding environment. The relationship between work and temperature is inverse, meaning that as work increases, temperature decreases and vice versa.

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