ganeshie8
  • ganeshie8
Suppose I have some gas confined in a cylinder with a movable piston as shown. I add some heat \(Q\) keeping the pressure constant. The volume changes by \(\Delta V\) and the temperature by \(\Delta T\). Next, keeping that volume constant, I remove the previously added heat \(Q\). What would be the change in temperature ? Options : 1) \(\Delta T\) 2) \(\gt \Delta T\) 3) \(\lt \Delta T\)
Mathematics
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
|dw:1450337283507:dw|
ganeshie8
  • ganeshie8
let me know if the question is ambiguous...
ParthKohli
  • ParthKohli
\[Q=\underbrace{p\Delta V}_{\text{as work}}+ \underbrace{nC_v \Delta T}_{\text{as internal energy}}\]Next, we remove \(Q\).\[-Q = -nC_v \Delta T'\]Well, this certainly seems to answer the question.

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ParthKohli
  • ParthKohli
In the second one, there is no work - only change in internal energy.
ganeshie8
  • ganeshie8
I think that shows the work done by the system(gas) is positive
ParthKohli
  • ParthKohli
Here, we see that the decrease in temperature is way more than what it increased by earlier.
ParthKohli
  • ParthKohli
\[nC_v \Delta T' = p\Delta V + nC_v \Delta T\]\[|\Delta T' | > |\Delta T|\]
ganeshie8
  • ganeshie8
that means \(\Delta T\) is not same in both the steps ?
ganeshie8
  • ganeshie8
Ohk gotcha !
ParthKohli
  • ParthKohli
No, it is not because we're using two different processes.
ganeshie8
  • ganeshie8
the quantity \(Q\) does not uniquely determine the change in temperature of one \(mol\) of gas then ?
ParthKohli
  • ParthKohli
In a constant-volume process, it does.
ganeshie8
  • ganeshie8
we're keeping pressure constant when adding heat and keeping volume constant when removing heat
Miracrown
  • Miracrown
\[PV = nRT\] http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html
Miracrown
  • Miracrown
There is also one more thing we need:' the 1st law of thermodynamics
ParthKohli
  • ParthKohli
oh nice, it also determines the constant-pressure\[p\Delta V = nR\Delta T \]
Miracrown
  • Miracrown
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html
ganeshie8
  • ganeshie8
|dw:1450337902286:dw|
ParthKohli
  • ParthKohli
thankfully here, we only needed the first law
ParthKohli
  • ParthKohli
\[Q = nR\Delta T + nC_v \Delta T\]Nice
Miracrown
  • Miracrown
When the pressure is kept constant, can we find the amount of work W?
ganeshie8
  • ganeshie8
Q - W = constant that is first law right
ParthKohli
  • ParthKohli
it's not constant it's the change in internal energy
Miracrown
  • Miracrown
Just one thing to keep in mind: Q - W = constant will NOT be true, because internal energy changes
ParthKohli
  • ParthKohli
you can look at it this way: \(Q = \Delta U + w\) Give a man some money (heat) - he'll spend some (work) and save the rest (internal energy).
Miracrown
  • Miracrown
It would be true in an isothermal (T=const) process
ganeshie8
  • ganeshie8
\[W = \int p dV = p\int dV \] |dw:1450338100977:dw|
Miracrown
  • Miracrown
but we don't have that, Our process is isobaric (P=const)
ganeshie8
  • ganeshie8
change in internal energy is constant between any two states that is first law right ?
ParthKohli
  • ParthKohli
yes, that is true. :)
ParthKohli
  • ParthKohli
but that's not the first law, that's saying that internal energy is a state function
ganeshie8
  • ganeshie8
Q - W between two states is constant no matter what the process is
Miracrown
  • Miracrown
It's one of the ways to state it @ganeshie8
Miracrown
  • Miracrown
Actually, it's one of the consequences of ideal gas law
ParthKohli
  • ParthKohli
yes ganeshie, but in this particular question, we're not actually going back to the same state
ParthKohli
  • ParthKohli
@Miracrown ideal gas law? it applies to any thermodynamical system
ganeshie8
  • ganeshie8
i think change in internal energy would be 0 if we go back to the same state
Miracrown
  • Miracrown
If you have state 1, with P1,V1,T1 and state 2, with P2,V2,T2 , then the change of U is uniquely determined
ParthKohli
  • ParthKohli
correct, but here we're not going back to the same state.
ParthKohli
  • ParthKohli
which is why you may have thought \(\Delta T_1 = \Delta T_2\)
ganeshie8
  • ganeshie8
I think thats why Q-W is constant, not 0
Miracrown
  • Miracrown
@ParthKohli - Well, yes! I was only thinking of ideal gas because that's the current problem we have
Miracrown
  • Miracrown
so we have said W=Fd
Miracrown
  • Miracrown
@ganeshie8 - That's true! the internal energy of ideal gas depends on T only
Miracrown
  • Miracrown
T1 and T2 (initial, final temp) determine delta U
Miracrown
  • Miracrown
so it will be a constant
ganeshie8
  • ganeshie8
\(|\Delta T_2| \gt |\Delta T_1|\)
Miracrown
  • Miracrown
can be computed as base area *height?
Miracrown
  • Miracrown
ΔV = Ad Let A be the piston area and d the distance it moved
ParthKohli
  • ParthKohli
\[dw = F dx\]\[= pAdx\]\[=p(Adx)\]\[=pdV\]
ganeshie8
  • ganeshie8
since p is constant, we can simply multiply dV by p yeah... after the process is finished, the temperature of gas would be less eventhohgh there is no net heat transfer
ganeshie8
  • ganeshie8
Oh, there is work done by the system... is that the reason for the reduced temperature ?
Miracrown
  • Miracrown
Q = F(pressure*area d = PAd = PΔV force, in its turn, is pressure*area so work is P*A*d which gives us P delta V. When P is constant, you can use W=P delta V formula
Miracrown
  • Miracrown
yes, we are told T changes
ParthKohli
  • ParthKohli
yeah ganeshie :)
Miracrown
  • Miracrown
That's part of the reason! @ganeshie8 Also, some of the added heat Q can be spent toward increasing U
Miracrown
  • Miracrown
Q = W + delta U and since W = P delta V we have here Q = P delta V + delta U
ganeshie8
  • ganeshie8
I see... since \(Q\) is 0 for the overall process (we have added Q and removed Q), the change in internal energy is because of work only : \(0 = W + \Delta U \implies \Delta U = -W\) Makes sense, thanks :)
Miracrown
  • Miracrown
You can find internal energy of the gas as U = (3/2) nRT (so just as I said, it depends only on temperature)
Miracrown
  • Miracrown
@ganeshie8 - Well, yes, but we're now only at the first stage!
ganeshie8
  • ganeshie8
Ohk... in the first stage, Q and W are both positive
Miracrown
  • Miracrown
Q = P delta V + 3/2 nR delta T
ganeshie8
  • ganeshie8
the gas is absorbing heat and also doing positive work
Miracrown
  • Miracrown
@ganeshie8 - Actually, we cannot be sure of that!
Miracrown
  • Miracrown
The problem only said that heat is ADDED so Q is positive
Miracrown
  • Miracrown
However, it did not tell us the sign of delta V
ganeshie8
  • ganeshie8
why not, since the pressure is constant, volume must increase right ?
ganeshie8
  • ganeshie8
By any chance, does that require using ideal gas law ?
ganeshie8
  • ganeshie8
because the textbook hasn't covered ideal gas law yet..
Miracrown
  • Miracrown
Maybe they compressed the gas, doing work on it (so gas will do negative work then) and letting T increase. In principle, it's also possible
Miracrown
  • Miracrown
Yes, you're right! We know that only because P=const Good observation!
ParthKohli
  • ParthKohli
yes, we don't need to know anything about gases quite yet
Miracrown
  • Miracrown
OK, maybe not, So far I have not used it But we do need U = 3/2 nRT
ParthKohli
  • ParthKohli
except U = C_v * T
ParthKohli
  • ParthKohli
yeah don't say 3/2 as it is not true for most gases
Miracrown
  • Miracrown
Correct, 3/2 is only for monotomic gas
ganeshie8
  • ganeshie8
that is true for monatomic gases for other gases we need consider spin and oscillations... fascinating stuff !
ganeshie8
  • ganeshie8
can't escape probability theory.. it keeps showing up in every subject...
Miracrown
  • Miracrown
I am still thinking whether this was implied by default. We'll change 3/2 if necessary
ganeshie8
  • ganeshie8
i think 3/2 is a safe assumption spin and oscillations require some minimum temperature... we may assume the gas is monatomic or the temperature requirement is not met for spin to start...
ganeshie8
  • ganeshie8
either case, that derivation is part of next chapter on kinetic theory of gases... we're good for now, thanks :)
ParthKohli
  • ParthKohli
oh but did you not calculate the RMS speed yesterday? it's only two steps away!
ganeshie8
  • ganeshie8
i have finished reading upto entropy and 2nd law doiing problems now..
Miracrown
  • Miracrown
Yes, I was just thinking that 3/2 should be assumed

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