Calculate the work (Thermodynamics)

[alivedude]

Homework Statement

An ideal gas with pressure $p_1=2$ bar and volume $V_1=3 \, m^3$ is expanding under constant pressure to the volume $V_2=3V_1$. Then we compress the gas isothermally back to $V_1$. Finally the gas cools down to initial pressure $p_1$. Calculate the total amount of work.

Homework Equations

Work done is given by $W=-\int_{V_1}^{V_2}pdV$

The Attempt at a Solution

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The first work is isobar so we have that
$W_1 = -p(V_2-V_1)$

For the last part we have that $dV=0$ so work done there is 0.

What about the second part? Work done for an isothermal compression is $W=-vRT\ln\frac{V_2}{V_1}$ but I have no information about the temperatur or number of mols. Can anyone guide me in the right direction?

The answer is $780$ kJ but I can't get there.

 

[stockzahn]

Hi,

$W=-vRT\ln\frac{V_2}{V_1}$

There is a mistake in this equation, this could be the problem.

 

[alivedude]

What is the mistake? I use the ideal gas law $pV=vRT$

$W=-vRT\int_{V_1}^{v_2} \frac{dV}{V} = -vRT \ln \frac{V_2}{V_1}$

 

[stockzahn]

$pV=vRT$

Maybe there is a misunderstanding in used letters: What dimension/value does $v$ have in your formula?

 

[alivedude]

Maybe there is a misunderstanding in used letters: What dimension/value does $v$ have in your formula?

Oh I'm sorry! It's in mol

 

[stockzahn]

OK, sorry too, I'm used to a different nomenclature. In your formulas

$pV=vRT$ & $W=-vRT\ln\frac{V_2}{V_1}$

you already have the solution. Just combine them. Hint: How does the product $p⋅V$ behave during an isothermal change of state?

 

[alivedude]

OK, sorry too, I'm used to a different nomenclature. In your formulas

$pV=vRT$ & $W=-vRT\ln\frac{V_2}{V_1}$

you already have the solution. Just combine them. Hint: How does the product $p⋅V$ behave during an isothermal change of state?

Im feeling bit slow here but i just can't see where i go wrong. I plug in $pV$ instead of $vRT$ and add the works

$W_{tot} = -p_1(V_2-V_1) - p_1V_2\ln \frac{V_2}{V_1}$

clearly this is wrong?

 

[stockzahn]

In your first post you already figured out, that for step (3) ΔW=0:

For the last part we have that $dV=0$ so work done there is 0.

To be able to reach the initial state again, there must be one step where the energy difference of the gas is positive and one step where it is negative. Your equation treats both steps like energy losses (work done by the gas).

Im feeling bit slow here but i just can't see where i go wrong. I plug in $pV$ instead of $vRT$ and add the works

$W_{tot} = -p_1(V_2-V_1) - p_1V_2\ln \frac{V_2}{V_1}$

clearly this is wrong?

 

[alivedude]

In your first post you already figured out, that for step (3) ΔW=0:
To be able to reach the initial state again, there must be one step where the energy difference of the gas is positive and one step where it is negative. Your equation treats both steps like energy losses.

To be honest, I have no idea what to do at this moment. The book i have is written by my professor and there is ZERO examples, just theory through the whole thing and I'm getting so frustrated. I have no idea what to do and I can't figure it out either.

This is what I'm thinking:

Since we reach the initial point the change in energy overall is zero. That means that the energy we add to the system by work is exactly the same as the amount of energy that we take away in form of heat when we cool the gas in step 3. During the cooling no work is done, so the only work comes from step 1 and 2. Then why isn't my calculation right here above?

 

[stockzahn]

Since we reach the initial point the change in energy overall is zero.

Correct

That means that the energy we add to the system by work is exactly the same as the amount of energy that we take away in form of heat when we cool the gas in step 3.

Not necessarily. You have to analyse each step seperatly to find out, during which step work is done by the gas (ΔW < 0) and during which step work is put into the gas (ΔW > 0). Have a look at the volume of the gas: Depending on its change, whether it is increasing or decreasing, work is done by the gas or put into the gas.

 

[alivedude]

Correct
Not necessarily. You have to analyse each step seperatly to find out, during which step work is done by the gas (ΔW < 0) and during which step work is put into the gas (ΔW > 0). Have a look at the volume of the gas: Depending on its change, whether it is increasing or decreasing, work is done by the gas or put into the gas.

Found out what the problem was and now i have the right answer.

My calculation for the work above was correct but I used $ln\frac{9}{3}$ instead of $ln\frac{3}{9}$. I wasn't paying attention that i was suppose to go back to $3 m^3$ again.

Anyway, thanks for all the patience you had with me!

 

[stockzahn]

You're welcome, I hope I could help.