Jump to content

1. Thermodynamic work


Scienc
 Share

Recommended Posts

What is the difference between the work done by a gas and the workflow that the same gas performs to exist? why does the variation in the internal energy of the equation dH = dQ + nRT not consider the work done by the gas to exist (working flow) but consumes only the heat transfer? 

In the question below, the author considers dU = Q. He performs the calculation of the work done by the gas, separated from the internal energy (with an equation dH = dU + nRT ). If you know that in systems with constant volume dU = Q, but the gas volume varies (From 6 moles to 12 moles of gas), this does not imply an internal energy variation equal to dU = w + q (As stated in the 1st law of thermodynamics) and why consider dU = w + q and not dU = Q **? In practice or in what he did, the logic he did not use was.

 

Question:

“A constant volume calorimeter showed that the fuel loss in burning 1 mole of glucose is equal to -2559kJ in 298K, that is dU = -2555KJ”

Link to comment
Share on other sites

4 minutes ago, Scienc said:

If you know that in systems with constant volume dU = Q, but the gas volume varies (From 6 moles to 12 moles of gas), this does not imply an internal energy variation equal to dU = w + q (As stated in the 1st law of thermodynamics) and why consider dU = w + q and not dU = Q **? In practice or in what he did, the logic he did not use was.

I think you're confusing change in volume with change in amount of matter present in your system. Moles are a measure of the number of atomic participants in your system. Volume is very different. Please review your statement so that a proper answer can be given to what's troubling you.

The system, such as you've defined it, is an open system (open to exchanges of both energy and matter).

Link to comment
Share on other sites

16 minutes ago, joigus said:

I think you're confusing change in volume with change in amount of matter present in your system. Moles are a measure of the number of atomic participants in your system. Volume is very different. Please review your statement so that a proper answer can be given to what's troubling you.

The system, such as you've defined it, is an open system (open to exchanges of both energy and matter).

The volume that 6 moles occupy 134.4l and 12 moles occupy 268.8 moles, since 1 mol = 22.4l, for this the gas needs to expand

Link to comment
Share on other sites

18 minutes ago, Scienc said:

The volume that 6 moles occupy 134.4l and 12 moles occupy 268.8 moles, since 1 mol = 22.4l, for this the gas needs to expand

I see. So it's a chemical reaction. Sorry. I should have read more carefully.

45 minutes ago, Scienc said:

and the workflow that the same gas performs to exist?

I really don't understand this part of one of your sentences. I don't have a concept of "workflow that a gas performs to exist."

Link to comment
Share on other sites

A constant volume calorimeter simply uses a constant volume to resist the increase in pressure due to the formation of combustion gasses from the original solid. The expansion of these gasses will cause the gasses to cool and therefore take away from the measured temperature change. Keeping the combustion gasses at a constant volume keeps the gasses from cooling via the Joule-Thompson effect and provides for a better measurement. The energy here is not coming from taking a solid and making it occupy a larger volume as a gas, the energy here comes from chemically changing the material by reacting it with oxygen. The measured quantity here is the enthalpy of reaction and has nothing to do with the potential for work done by the expansion of the gasses.

Edited by AllMyFriendsAreDangerous
Link to comment
Share on other sites

1 hour ago, Scienc said:

What is the difference between the work done by a gas and the workflow that the same gas performs to exist? why does the variation in the internal energy of the equation dH = dQ + nRT not consider the work done by the gas to exist (working flow) but consumes only the heat transfer? 

Most chemical reactions of interest take place at constant pressure. Irrespective of how general that assumption is, most chemistry books do consider P to be constant. If that's the case, even though partial pressures of the different components are changing:

\[p_{i}=n_{i}\frac{RT}{V}\]

because the moles are changing. And assuming the changes occur in quasi-static conditions, while every ni is changing, the total volume is not. Nor is the total pressure. So the total work is zero if the volume doesn't change.

Maybe that has to do with what you mean by "the workflow that the gas performs to exist". Is that it? Are you picturing that the gas that's being formed does work in order to be formed? That's not the case if the volume is also constant. Mind you, there is no such a thing as "partial volumes".

I hope that answers your question.

Edited by joigus
minor addition
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.