Question

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) \(\lt \Delta T\)

Answer 1

let me know if the question is ambiguous...

Answer 2

\[Q=\underbrace{p\Delta V}_{\text{as work}}+ \underbrace{nC_v \Delta T}_{\text{as internal energy}}\]Next, we remove \(Q\).\[-Q = -nC_v \Delta T'\]Well, this certainly seems to answer the question.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Ohk gotcha !

Answer 9

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

Answer 10

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

Answer 11

In a constant-volume process, it does.

Answer 12

we're keeping pressure constant when adding heat

and keeping volume constant when removing heat

Answer 13

\[PV = nRT\]

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html

Answer 14

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

Answer 15

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

Answer 16

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