Thermodynamics - Work done per unit mass

  • Thread starter davcrai
  • Start date
  • #1
davcrai
13
0

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???
 
Last edited:

Answers and Replies

  • #2
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,134
161


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
davcrai
13
0


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
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,134
161


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
davcrai
13
0


Hmmm, might it include density and a gas constant by any chance....
Thanks
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,134
161


It would definitely include the gas constant.
 

Suggested for: Thermodynamics - Work done per unit mass

Replies
0
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
19K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
951
Replies
6
Views
2K
Replies
2
Views
4K
Top