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Thermodynamics - Work done per unit mass

  • Thread starter davcrai
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Homework Statement



A sample of dry air has initial pressure p1 = 1000 hPa and temperature T1 = 300k. It undergoes a process that takes it to a new pressure p2 = 500 hPa with unchanged temperature T2 = T1. Compute the mechanical work per unit mass performed by the sample under the following scenarios:
a) Isochoric pressure reduction to p2 followed by isobaric expansion to final state.
b) Isobaric expansion to final specific volume s2 followed by isochoric pressure reduction to final state.
c) Isothermal expansion to final state.


Homework Equations



work = ∫pds
s = Volume/mass


The Attempt at a Solution



a) I think pressure held constant at 500 and the integral preformed between s1 and s2, to get -500*(s2-s1).

b) The integral is ∫p(s)ds with limits s1 to s2 from to give p(s)*(s2-s1), then evaluate at p = 500 for the same result ???

c) The previous two were isothermal???
 
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Answers and Replies

  • #2
Redbelly98
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Homework Statement



A sample of dry air has initial pressure p1 = 1000 hPa and temperature T1 = 300k. It undergoes a process that takes it to a new pressure p2 = 500 hPa with unchanged temperature T2 = T1. Compute the mechanical work per unit mass performed by the sample under the following scenarios:
a) Isochoric pressure reduction to p2 followed by isobaric expansion to final state.
b) Isobaric expansion to final specific volume s2 followed by isochoric pressure reduction to final state.
c) Isothermal expansion to final state.


Homework Equations



work = ∫pds
s = Volume/mass


The Attempt at a Solution



a) I think pressure held constant at 500 and the integral preformed between s1 and s2, to get -500*(s2-s1).
Looks reasonable so far, but you need to figure out values for s1 and s2. Also, watch the units.

b) The integral is ∫p(s)ds with limits s1 to s2 from to give p(s)*(s2-s1), then evaluate at p = 500 for the same result ???
The pressure is not 500 hPa for this one; the first step is isobaric in this case. Try drawing yourself a PV diagram, to help figure out what is going on.

c) The previous two were isothermal???
Um, no, they were not. There were two steps, one isochoric and one isobaric. So the temperature changed during processes (a) and (b), even though it ended up at the initial temperature.

This time, it is isothermal during the entire process.
 
  • #3
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ok, so I think for the second one you hold p constant at 1000hPa for the integration then evaluate between s1 and s2??????
Still unsure about the third one????
 
  • #4
Redbelly98
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ok, so I think for the second one you hold p constant at 1000hPa for the integration then evaluate between s1 and s2??????
Yes. You'll also need to figure out what s1 and s2 are.
Still unsure about the third one????
You need to find an equation that relates P and s, using the fact that T=constant. (Hint: the substance is air, and air is a gas.)

Once you have the relation between P and s, do the integral ∫p·ds.
 
  • #5
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Hmmm, might it include density and a gas constant by any chance....
Thanks
 
  • #6
Redbelly98
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It would definitely include the gas constant.
 

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