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) \(\lt \Delta T\)

- ganeshie8

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- ganeshie8

|dw:1450337283507:dw|

- ganeshie8

let me know if the question is ambiguous...

- 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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## More answers

- ParthKohli

In the second one, there is no work - only change in internal energy.

- ganeshie8

I think that shows the work done by the system(gas) is positive

- ParthKohli

Here, we see that the decrease in temperature is way more than what it increased by earlier.

- ParthKohli

\[nC_v \Delta T' = p\Delta V + nC_v \Delta T\]\[|\Delta T' | > |\Delta T|\]

- ganeshie8

that means \(\Delta T\) is not same in both the steps ?

- ganeshie8

Ohk gotcha !

- ParthKohli

No, it is not because we're using two different processes.

- ganeshie8

the quantity \(Q\) does not uniquely determine the change in temperature of one \(mol\) of gas then ?

- ParthKohli

In a constant-volume process, it does.

- ganeshie8

we're keeping pressure constant when adding heat
and keeping volume constant when removing heat

- Miracrown

\[PV = nRT\]
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html

- Miracrown

There is also one more thing we need:'
the 1st law of thermodynamics

- ParthKohli

oh nice, it also determines the constant-pressure\[p\Delta V = nR\Delta T \]

- Miracrown

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html

- ganeshie8

|dw:1450337902286:dw|

- ParthKohli

thankfully here, we only needed the first law

- ParthKohli

\[Q = nR\Delta T + nC_v \Delta T\]Nice

- Miracrown

When the pressure is kept constant, can we find the amount of work W?

- ganeshie8

Q - W = constant
that is first law right

- ParthKohli

it's not constant
it's the change in internal energy

- Miracrown

Just one thing to keep in mind: Q - W = constant will NOT be true, because internal energy changes

- 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

It would be true in an isothermal (T=const) process

- ganeshie8

\[W = \int p dV = p\int dV \]
|dw:1450338100977:dw|

- Miracrown

but we don't have that, Our process is isobaric (P=const)

- ganeshie8

change in internal energy is constant between any two states
that is first law right ?

- ParthKohli

yes, that is true. :)

- ParthKohli

but that's not the first law, that's saying that internal energy is a state function

- ganeshie8

Q - W between two states is constant no matter what the process is

- Miracrown

It's one of the ways to state it @ganeshie8

- Miracrown

Actually, it's one of the consequences of ideal gas law

- ParthKohli

yes ganeshie, but in this particular question, we're not actually going back to the same state

- ParthKohli

@Miracrown ideal gas law? it applies to any thermodynamical system

- ganeshie8

i think change in internal energy would be 0 if we go back to the same state

- 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

correct, but here we're not going back to the same state.

- ParthKohli

which is why you may have thought \(\Delta T_1 = \Delta T_2\)

- ganeshie8

I think thats why Q-W is constant, not 0

- Miracrown

@ParthKohli - Well, yes! I was only thinking of ideal gas because that's the current problem we have

- Miracrown

so we have said W=Fd

- Miracrown

@ganeshie8 - That's true! the internal energy of ideal gas depends on T only

- Miracrown

T1 and T2 (initial, final temp) determine delta U

- Miracrown

so it will be a constant

- ganeshie8

\(|\Delta T_2| \gt |\Delta T_1|\)

- Miracrown

can be computed as base area *height?

- Miracrown

ΔV = Ad
Let A be the piston area and d the distance it moved

- ParthKohli

\[dw = F dx\]\[= pAdx\]\[=p(Adx)\]\[=pdV\]

- 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

Oh, there is work done by the system... is that the reason for the reduced temperature ?

- 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

yes, we are told T changes

- ParthKohli

yeah ganeshie :)

- Miracrown

That's part of the reason! @ganeshie8
Also, some of the added heat Q can be spent toward increasing U

- Miracrown

Q = W + delta U and since W = P delta V
we have here Q = P delta V + delta U

- 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

You can find internal energy of the gas as
U = (3/2) nRT
(so just as I said, it depends only on temperature)

- Miracrown

@ganeshie8 - Well, yes, but we're now only at the first stage!

- ganeshie8

Ohk... in the first stage, Q and W are both positive

- Miracrown

Q = P delta V + 3/2 nR delta T

- ganeshie8

the gas is absorbing heat and also doing positive work

- Miracrown

@ganeshie8 - Actually, we cannot be sure of that!

- Miracrown

The problem only said that heat is ADDED so Q is positive

- Miracrown

However, it did not tell us the sign of delta V

- ganeshie8

why not, since the pressure is constant, volume must increase right ?

- ganeshie8

By any chance, does that require using ideal gas law ?

- ganeshie8

because the textbook hasn't covered ideal gas law yet..

- 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

Yes, you're right! We know that only because P=const
Good observation!

- ParthKohli

yes, we don't need to know anything about gases quite yet

- Miracrown

OK, maybe not, So far I have not used it But we do need U = 3/2 nRT

- ParthKohli

except U = C_v * T

- ParthKohli

yeah
don't say 3/2 as it is not true for most gases

- Miracrown

Correct, 3/2 is only for monotomic gas

- ganeshie8

that is true for monatomic gases
for other gases we need consider spin and oscillations... fascinating stuff !

- ganeshie8

can't escape probability theory.. it keeps showing up in every subject...

- Miracrown

I am still thinking whether this was implied by default. We'll change 3/2 if necessary

- 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

either case, that derivation is part of next chapter on kinetic theory of gases... we're good for now, thanks :)

- ParthKohli

oh but did you not calculate the RMS speed yesterday? it's only two steps away!

- ganeshie8

i have finished reading upto entropy and 2nd law
doiing problems now..

- Miracrown

Yes, I was just thinking that 3/2 should be assumed

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