Thermodynamics 1st law problem

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Homework Statement


Air occupies [itex]0.140m^3[/itex] while under gauge pressure of [itex]1.03x10^5[/itex]

It is then expanded isothermically to atmospheric pressure.

Then it is cooled at constant preesure back to its initial volume.

And then I should calculate the work done by the gas.

Homework Equations



[itex] W=(nRT) \ln \frac{v_{1}}{v_{2}} [/itex]
[itex] PV=nRT [/itex]

The Attempt at a Solution



First I need to know how many mols of air I have.
So I convert [itex] 0.140 m^3 [/itex] to centimeters and since 1 mol of anything occupies [itex]22.4 L [/itex] or [itex]22400 cm^3[/itex] I find that I have 6.25mols of air.


Then I find the initial temperature of the system using [itex] T_{i} = \frac{PV}{Rn} = 277.6K [/itex]

And then I'm stuck here

I tried before calculating the final volume and with this the ln between initial and final volumes.
but I made the wrong assumption that temperature was constant which is not true.

Now I'm stuck because I'm not sure how to calculate the final volume.
 

Answers and Replies

  • #2
Simon Bridge
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Hint: Draw the P-V diagram.
The work done is the area inside the loop.

Note: 1mol of a gas can have different volumes for different pressures and temperatures.
You don't need to know the number of moles or the temperature: you need to know (nRT)
 
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  • #3
Simon Bridge
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System starts at state A, goes to state B, then state C.

A: PAVA=nRT1 ...(1)
B: PBVB=nRT1 ...(2)
C: PBVA=nRT2 ...(3)

know: PA, VA, PB
don't know: VB, (nRT1), and (nRT2)

Work done is the area under the P-V diagram.
From the diagram:

[itex]W=(nRT_1)\ln |\frac{V_B}{V_A}| - P_B(V_B - V_A)[/itex] ...(4)

from (1) ... [itex](nRT_1)=P_AV_A[/itex] ...(5)

divide (1) into (2)

[itex]V_B =P_AV_A/P_B[/itex] ...(6)

substitute (5) and (6) into (4) and simplify:

[itex]W=(P_AV_A)\ln |\frac{P_A}{P_B}| - V_A(P_A - P_B)[/itex]

[edit]<sigh> didn't mean to post it yet! oh well... make sure you understand what I did.
 
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  • #4
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You don't need to know the number of moles or the temperature: you need to know (nRT)
I honestly haven't thought about that...


Anyways: the answer in the book is 5 700 N m (for some reason it didnt use Joules)
Unless I miscalculated somewhere, trying your approach gives me a result of 2.75J
When I tried another method it gave me 282.7 J
 
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  • #5
Simon Bridge
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1N.m = 1J

Did you remember to add atmospheric pressure to the gauge-pressure for PA?
Check your units are consistent.

Note: the method I show has to be correct: these are the three state equations, and there are three unknowns, so the system is known completely and unambiguously.
You do need to check my derivation... I may have misplaced a minus sign or something.

You left off the units for pressure in post #1 so I cannot check.
[itex]1.03 \times 10^5[/itex] what?
 
  • #6
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1N.m = 1J
I know, it just kinda feels awkward

Did you remember to add atmospheric pressure to the gauge-pressure for PA?
Nope.
That is probably what is messing things up.
edit: tried adding, still not solving.
I keep getting 1.3J as answer

You left off the units for pressure in post #1 so I cannot check.
[itex]1.03 \times 10^5[/itex] what?
Pa
 
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  • #7
Simon Bridge
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OK: I get -

PA = 204kPa
VA = 140L
nRT1 = 29172kPa.L
PB = 101kPa
VB = 29172/101 = 289L
nRT2 = 29172x140/289 = 14132kPa.L [*]

WAB = 29172xln|289/140| = 21143kPa.L (isothermal)
WBC = -101x(289-140) = -15049kPa.L (isobaric)
WCA = 0 (isochoric) [*]

WTOT = 21143 - 15049 = 6094kPa.L

note:
1kPa.L = 1000(N/m2)L and 1000L=1m3 so 1kPa.L=1N.m

It's in the right ballpark - maybe less rounding off will get the 5700Nm needed?
I cannot tell how you got such low figures without seeing your calculations.

[*] these entries included for completeness.
 
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  • #8
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probably because I was using cubic meters instead of liters.
giving

PA = 204kPa
VA = 0.140 m^3
and then
nRT1 = 28560Pa.m^3

PB = 101kPa
VB = 28560Pa/101kPa =0. 283m^3
nRT2 = 28560x0.140/0.283 = 14129Pa


WAB = 28560Pa x ln|0.283/0.140| = 20100Pa.m^3 (isothermal)
WBC = -101kPa x(0.283 -0.140) = -14443Pa.L (isobaric)
giving WTOT = 20100 - 14443 = 5657 J


The answer in the book is 5700 J
and the difference is probably because of rounding figures.

Well, this time I got it right.
But I'm not sure what was I doing that gave me those small numbers...


Thanks for the help
 
  • #9
Simon Bridge
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giving WTOT = 20100 - 14443 = 5657 J

The answer in the book is 5700 J
and the difference is probably because of rounding figures.
Well done!

The difference is actually because you didn't do enough rounding:

The states are all known to 3 sig fig, so you should give your answer to 3 sig fig: 5657J is 4 sig fig - rounding to three gives you 5700J.

When you give a formal answer you want to have:

"Wtot = 5657J = 5700J (3sig.fig.)"

as the 2nd to last line (last line is the same thing in words).

aside: I just miskeyed a number into the computer :)
What I've just modeled for you is a disciplined way to carry out analysis and calculations ... see how just listing everything like that clears a lot of thinking.
Also very important is to draw the graph - most of your thermodynamics is is encoded in that and realizing that the work in any stage is the area under the graph.
Even if you remember the equations, looking at the graph clears things up.
 
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