Heat and work when temperature increases by 1 degree

[Helly123]

Homework Statement

Kinetic energy per mol is 3/2KT

Homework Equations

Q = nC$\Delta$T
U = Q + W
W = -P$\Delta$V

The Attempt at a Solution

1) internal energy = 3/2NKT

2) heat needed to increase temperature of 1 mol ideal gas by 1 degree at constant volume?
Since constant volume, W = 0
Q = Internal change
Q = 3/2NKT= 3/2nRT

3) work done by one mole ideal gas, when increased by 1 degree temperature, constant pressure?
W done = Q - U

4) when volume doubled, how much increased in kinetic energy?
I thought that more volume means less kinetic energy

5) volume doubled, adiabatic condition, final pressure must 0.3 lower than first pressure.

What i focused is on numbers 2 and 3
Can i get a clue?

 

[mjc123]

The question is impossible to answer if you don't know whether it is a monatomic, diatomic or polyatomic gas - i.e. how many internal degrees of freedom there are. The average translational kinetic energy is (3/2)NkT (note we use a small k for Boltzmann's constant), but to find the internal energy we add (1/2)NkT per rotational degree of freedom, and NkT per active vibrational mode. (Strictly I should say we add 1/2Nk and Nk respectively to the heat capacity; the internal energy will be less than stated because the heat capacity is lower at very low temperatures.)

Be careful to distinguish quantities from changes in these quantities. In 2), Q is not 3/2nRT, but 3/2nRΔT (for a monatomic gas).

when volume doubled, how much increased in kinetic energy?

Can't answer if it doesn't say under what conditions, e.g. halve the pressure at constant T, or double temperature at constant P?

 

[Helly123]

The question is impossible to answer if you don't know whether it is a monatomic, diatomic or polyatomic gas - i.e. how many internal degrees of freedom there are. The average translational kinetic energy is (3/2)NkT (note we use a small k for Boltzmann's constant), but to find the internal energy we add (1/2)NkT per rotational degree of freedom, and NkT per active vibrational mode. (Strictly I should say we add 1/2Nk and Nk respectively to the heat capacity; the internal energy will be less than stated because the heat capacity is lower at very low temperatures.)

Be careful to distinguish quantities from changes in these quantities. In 2), Q is not 3/2nRT, but 3/2nRΔT (for a monatomic gas).Can't answer if it doesn't say under what conditions, e.g. halve the pressure at constant T, or double temperature at constant P?

It wasn't stated what gas. I guess it is monoatomic generally.
For 4) it is at constant temperature

 

[mjc123]

If it's at constant temperature, what does that imply about the kinetic energy, given the first equation you quote?

 

[Helly123]

Ok.

If it's at constant temperature, what does that imply about the kinetic energy, given the first equation you quote?

Kinetic energy is 3/2Nk$\Delta$T. It will only depend on N. For T is constant. N is depend on volume. It is inversely propotional?

 

[Helly123]

How about 2) ? Is heat (Q) = internal change? For W = 0. The answer is not 3/2nR$\Delta$T. The answer covers NkT formula, instead. So, N is number of molecules (still try to understand the difference between N for molecules and n for moles)

 

[mjc123]

N is the number of molecules, not the number density. I assume we're talking about a constant amount of gas, subjected to P, V or T changes. So N is constant.

For constant volume, W = 0, so Q = ΔU (not U. Do you see why?). What is the value of n? What is ΔT? What is N for 1 mole of gas?

 

[Helly123]

N is the number of molecules, not the number density. I assume we're talking about a constant amount of gas, subjected to P, V or T changes. So N is constant.

For constant volume, W = 0, so Q = ΔU (not U. Do you see why?). What is the value of n? What is ΔT? What is N for 1 mole of gas?

N for 1 mole gas is 6.03 x 10^23 /moles* 1 moles
How can we deduce that N is constant

 

[Helly123]

N is the number of molecules, not the number density. I assume we're talking about a constant amount of gas, subjected to P, V or T changes. So N is constant.

For constant volume, W = 0, so Q = ΔU (not U. Do you see why?). What is the value of n? What is ΔT? What is N for 1 mole of gas?

Q = ΔU
Because T changes, U changes, so Q changes.

Q = U final - U initial
Q = 3/2Nk(Tf - Ti)
Q = 3/2Nk(Ti + 1 - Ti)
Q = 3/2Nk

Is that it?

But i am not sure, because U final should be the same as U initial. Internal change always the same, right?

 

[Helly123]

Or maybe W = -PΔV
V1/T1 = V2/T2
T2 = (T1 + 1)

 

[mjc123]

N for 1 mole gas is 6.03 x 10^23 /moles* 1 moles
How can we deduce that N is constant

Because, as I said, we are assuming that we are dealing with a fixed amount of gas (and "amount" means number of moles, not volume) undergoing various changes, without adding or removing any gas. If that is not the case, the question should tell you so.

Q = ΔU
Because T changes, U changes, so Q changes.

Be careful here. Q is a definite, measurable quantity, the amount of heat that is gained or lost by the system. We give this quantity the symbol Q. It is related to changes in state functions of the system, such as U. This change we call ΔU. U itself is not absolutely defined - there is no absolute zero of energy; all energies are relative to some more or less arbitrarily defined zero - but ΔU is well defined and measurable.
That is why we write Q = ΔU, with a delta before U but not Q. ΔU is a change, but Q is simply a measured quantity. In a simple case like this, we are not concerned with any change in Q. There are cases in which we may be - e.g. we might ask "What is the difference in Q if we do this process under different conditions (e.g. constant pressure vs. constant volume)?"

But i am not sure, because U final should be the same as U initial. Internal change always the same, right?

Why? You raise the temperature, the internal energy rises. Is that not obvious? Is that not what a change in temperature means?

 

[Helly123]

Because, as I said, we are assuming that we are dealing with a fixed amount of gas (and "amount" means number of moles, not volume) undergoing various changes, without adding or removing any gas. If that is not the case, the question should tell you so.

Be careful here. Q is a definite, measurable quantity, the amount of heat that is gained or lost by the system. We give this quantity the symbol Q. It is related to changes in state functions of the system, such as U. This change we call ΔU. U itself is not absolutely defined - there is no absolute zero of energy; all energies are relative to some more or less arbitrarily defined zero - but ΔU is well defined and measurable.
That is why we write Q = ΔU, with a delta before U but not Q. ΔU is a change, but Q is simply a measured quantity. In a simple case like this, we are not concerned with any change in Q. There are cases in which we may be - e.g. we might ask "What is the difference in Q if we do this process under different conditions (e.g. constant pressure vs. constant volume)?"

Why? You raise the temperature, the internal energy rises. Is that not obvious? Is that not what a change in temperature means?

How about when there is a graph of PV from point A-B-C. Then $\Delta$U at A = B = C

 

[Helly123]

Q = U final - U initial
Q = 3/2Nk(Tf - Ti)
Q = 3/2Nk(Ti + 1 - Ti)
Q = 3/2Nk

so, is this true?

 

[mjc123]

How about when there is a graph of PV from point A-B-C. Then ΔΔ\DeltaU at A = B = C

Where are points A, B and C? Are they points on a PV isotherm (i.e. T is constant and PV is constant)? Then U is the same at A, B and C and ΔU = 0 for the process. (Note the difference between U and ΔU, you don't seem to have got it yet.)

so, is this true?

You have to define N for 1 mole. As Q is a macroscopic quantity, I would prefer to say Q = 3/2 R.

 

[Helly123]

Where are points A, B and C? Are they points on a PV isotherm (i.e. T is constant and PV is constant)? Then U is the same at A, B and C and ΔU = 0 for the process. (Note the difference between U and ΔU, you don't seem to have got it yet.)

You have to define N for 1 mole. As Q is a macroscopic quantity, I would prefer to say Q = 3/2 R.

Ok.
How about the work done when temperature increased by 1 degree?
The increase of T will affect the Volume.
So there is work. T increase, V2 will increase. Gas does work (work negative).
W = - (-$\int$P$\Delta$V) = P$\Delta$V = NK$\Delta$T = NK((Ti + 1) - Ti) = Nk
Is that it?

 

[mjc123]

Yes, though again I would say R.
Oh, and don't put a capital Δ in an integral. Work = ∫PdV = PΔV at constant pressure.

 

[Helly123]

Yes, though again I would say R.
Oh, and don't put a capital Δ in an integral. Work = ∫PdV = PΔV at constant pressure.

I see. Ok thanks

 

[Helly123]

Btw.. when volume doubled at constant temperature, will Ek change?
Since Ek = 3/2nR$\Delta$T
I think it is not changed

 

[mjc123]

What is the relationship between Ek and temperature? (Hint: it does not involve something called ΔT. Seriously, learn the difference between T and ΔT etc. You seem to be throwing deltas around without understanding what they mean.) Do you think Ek should change at constant T?

 

[Helly123]

What is the relationship between Ek and temperature? (Hint: it does not involve something called ΔT. Seriously, learn the difference between T and ΔT etc. You seem to be throwing deltas around without understanding what they mean.) Do you think Ek should change at constant T?

I still get problen. Sorry.
But i think Ek not change at constant T.
The question say, volume doubled at constant T, how much Ek change

 

[mjc123]

You are correct, Ek does not change at constant temperature.

 

[Helly123]

You are correct, Ek does not change at constant temperature.

Ok. But the the answer is once greater. The Ek increase. It said that volume doubled at constant temperature.

For 2nd problem. At adiabatic process, P = 0.3 lower than P initial. Volume doubled. How much change in Ek?

I thought that T2 = (P2V2/P1V1 )T1
T2 = 2V*0.7P/V*P T1 = 1.4 T1

Ek 2 = 3/2nR(1.4T) = 0.4 greater than Ek1. But still wrong. Please help me