What is u in the equation for delta U=m(u2-u1)?

[EastWindBreaks]

Homework Statement

Homework Equations

The Attempt at a Solution

U is internal energy, but what is u? if its internal energy at a specific state, then why its multiplied by mass? does not makes sense to me...

 

[mfb]

Looks like the specific energy. More context would help.

 

[EastWindBreaks]

Looks like the specific energy. More context would help.

sorry, the question is : A piston-cylinder device contains 0.95 kg of oxygen initially at a temperature of 27 Celsius and a pressure, due to the ambient atmospheric pressure and a weight on the top of the piston, of 150 kPa. Heat is added to the gas until it reaches a temperature of 627 Celsius . Determine the amount of heat added to the gas during the process.

Q-W=Delta U
how does delta E= delta U= m(u2-u1)=m*c*(T2-T1)?delta U = Q is only correct under constant volume, for which has no work. so, since this thing is changing its volume, what happened to the work done by the system, its not in the equation. (this is from the book),as far as I know, m*c*(T2-T1) is just Q, heat energy.

and yeah, u as specific energy makes sense to me...just first time seeing specific energy. Thank You!

 

[whitsona]

Can you say more about "this is from the book"? Does the book say that you will use only this formula to figure out the answer? Or, does the book say that no work is done by the gas? I mean, does the book say, "the volume did not change". Or, is it asking you to calculate the change in volume (and therefore the work done) as a step in this problem?

 

[whitsona]

Found an answer for you. Sorry to refer you to Wikipedia. But, their page on isobaric processes derives this for you. Note that you are using the OTHER specific heat. :-) There is cv and there is cp. I'm sorry to be so rusty! Too many years in engineering management. Sad, really.

 

[whitsona]

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

 

[Chestermiller]

Homework Statement

Homework Equations

The Attempt at a Solution

U is internal energy, but what is u? if its internal energy at a specific state, then why its multiplied by mass? does not makes sense to me...

u is the internal energy per unit mass.

 

[Chestermiller]

sorry, the question is : A piston-cylinder device contains 0.95 kg of oxygen initially at a temperature of 27 Celsius and a pressure, due to the ambient atmospheric pressure and a weight on the top of the piston, of 150 kPa. Heat is added to the gas until it reaches a temperature of 627 Celsius . Determine the amount of heat added to the gas during the process.

Q-W=Delta U
how does delta E= delta U= m(u2-u1)=m*c*(T2-T1)?delta U = Q is only correct under constant volume, for which has no work. so, since this thing is changing its volume, what happened to the work done by the system, its not in the equation. (this is from the book),as far as I know, m*c*(T2-T1) is just Q, heat energy.

and yeah, u as specific energy makes sense to me...just first time seeing specific energy. Thank You!

For an ideal gas, the internal energy is a function only of temperature (and not volume). So it doesn't matter whether the volume changed. The subscript v on Cv refers to how this quantity is measured experimentally in the laboratory (by determining the amount of heat Q needed to raise the temperature of a material at constant volume, i.e., in a test where ##Q =\Delta U=mc\Delta T##), and not how Cv is applied in practice for more general situations (where ##Q\neq\Delta U##).

 

[EastWindBreaks]

Found an answer for you. Sorry to refer you to Wikipedia. But, their page on isobaric processes derives this for you. Note that you are using the OTHER specific heat. :-) There is cv and there is cp. I'm sorry to be so rusty! Too many years in engineering management. Sad, really.

For an ideal gas, the internal energy is a function only of temperature (and not volume). So it doesn't matter whether the volume changed. The subscript v on Cv refers to how this quantity is measured experimentally in the laboratory (by determining the amount of heat Q needed to raise the temperature of a material at constant volume, i.e., in a test where ##Q =\Delta U=mc\Delta T##), and not how Cv is applied in practice for more general situations (where ##Q\neq\Delta U##).

sorry, I am still a little confused, are you saying that for an ideal gas, Delta U simply equals Q? but if the volume changes as it is for this problem, the work done does not contribute to the internal energy of the system? that doesn't makes sense to me, if the work done doesn't contribute to the internal energy in this case, how its going to contribute in other cases? i thought work can raise temperature as well?

On Wikipedia it states " assuming that the quantity of gas stays constant, e.g., there is no phase transition during a chemical reaction. According to the equipartition theorem, the change in internal energy is related to the temperature of the system by

,
where cV is specific heat at a constant volume."

I have never heard of equipartition theorem, I looked into its page, but didn't find an explanation for how delta U is just Q where work is not considered.

 

[Chestermiller]

sorry, I am still a little confused, are you saying that for an ideal gas, Delta U simply equals Q?

No. Delta U is equal to Q only if no work is done. But, for an ideal gas, ##\Delta U=nC_v\Delta T## always. So, for an ideal gas, ##Q=nC_v\Delta T## only if no work is done. You must learn to associate Cv with U, and must stop associating Cv with Q. Otherwise, this is going to drive you crazy.

but if the volume changes as it is for this problem, the work done does not contribute to the internal energy of the system?

The work contributes to internal energy through the first law equation ##\Delta U=mC_v\Delta T=Q-W##. Note that in this equation, if work is done, the internal energy is affected.

i thought work can raise temperature as well?

Yes, in the sense I just mentioned.

On Wikipedia it states " assuming that the quantity of gas stays constant, e.g., there is no phase transition during a chemical reaction. According to the equipartition theorem, the change in internal energy is related to the temperature of the system by

,
where cV is specific heat at a constant volume."

I have never heard of equipartition theorem, I looked into its page, but didn't find an explanation for how delta U is just Q where work is not considered.

This equation is always correct for the case of an ideal gas. It can be regarded as the definition of Cv. Back to the first law. If W = 0, then ##\Delta U=Q##. This is the only circumstance in which we also find that also ##Q=nC_v\Delta T##.

 

[Chestermiller]

In summary, for an ideal gas,

1. ##\Delta U## is always equal to ##nC_v\Delta T##

2. Q is equal to ##nC_v\Delta T## only if no work is done

 

[EastWindBreaks]

In summary, for an ideal gas,

1. ##\Delta U## is always equal to ##nC_v\Delta T##

2. Q is equal to ##nC_v\Delta T## only if no work is done

I see, constant volume = no work, that makes sense that ΔU=mCvΔT. But for this problem, one of the step in the book wrote ΔU=mCvΔT, the volume does change though...the pressure is constant.

 

[Chestermiller]

I see, constant volume = no work, that makes sense that ΔU=mCvΔT. But for this problem, one of the step in the book wrote ΔU=mCvΔT, the volume does change though...the pressure is constant.

For an ideal gas, internal energy is a function only of temperature. So, no, constant volume is not required in order for ##\Delta U=nC_v\Delta T##. Like I said(several times), this equation applies irrespective of whether volume is constant. I don’t know how many different ways I can say this.

 

[EastWindBreaks]

For an ideal gas, internal energy is a function only of temperature. So, no, constant volume is not required in order for ##\Delta U=nC_v\Delta T##. Like I said(several times), this equation applies irrespective of whether volume is constant. I don’t know how many different ways I can say this.

but it was ##\Delta U=mC_v\Delta T## in the book, m not n

 

[Chestermiller]

but it was ##\Delta U=mC_v\Delta T## in the book, m not n

Whichever you prefer.

 

[EastWindBreaks]

Whichever you prefer.

m and n are completely different though, m for mass, n for number of moles,

 

[Chestermiller]

m and n are completely different though, m for mass, n for number of moles,

So, if you use m, then you use the heat capacity per unit mass, and if you use n, then you use the heat capacity per mole.

 

[EastWindBreaks]

So, if you use m, then you use the heat capacity per unit mass, and if you use n, then you use the heat capacity per mole.

I finally got it, with molar heat capacity being capital letter C and specific heat per unit mass being lower case c, I didn't know there are 2 version of heat capacities, thank you!