Another Thermo Question

  • Thread starter suspenc3
  • Start date
  • #1
suspenc3
402
0

Homework Statement



See Attachment



The Attempt at a Solution




Well to start off I know the gas is monatomic

I can find the work done by finding the area under each curve/line:

[tex]W_{AB} = p_0(2v_0-v_0) = p_0v_0[/tex]
[tex]W_{CD} = (p_0/32)(8v_0-16v_0) = \frac{-p_0v_0}{4}[/tex]

I can't think of how to find the area under the other two curves, I am guessing integration but I don't know how to set it up.

After I find these other to, what do I do?

Thanks
 

Attachments

  • phys2.jpg
    phys2.jpg
    24.5 KB · Views: 343

Answers and Replies

  • #2
FredGarvin
Science Advisor
5,088
10
I can't see the attachment yet, but finding the area between two curves is easy. The easiest way to do it is to subtract the area under the lower curve from the area under the upper curve.

If you did it in one integral, you could set up the limits of integration from one curve to the other.
 
  • #4
eaboujaoudeh
150
0
for isentropic expansion and compression in ICEs we use the same principle u used to find the first 2 works.
we use : W= (PoVo-PoVo/4)/(1-k) k=Cp/Cv. for isentropic adiabatic compression or expansion.

How did u know the gas was monoatomic?
 
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,740
438
for isentropic expansion and compression in ICEs we use the same principle u used to find the first 2 works.
we use : W= (PoVo-PoVo/4)/(1-k) k=Cp/Cv. for isentropic adiabatic compression or expansion.
Where do you get this? It works for BC but not DA

Generally, for reversible adiabatic paths:

(1) [tex]W = K\frac{V_f^{1-\gamma} - V_i^{1-\gamma}}{1-\gamma}[/tex]

where [itex]K = PV^\gamma[/itex]

This is just the integral [itex]\int dW[/itex] where [itex]dW = dU = PdV = KV^{-\gamma}dV [/itex] (dQ=0)

Since for DA [itex]P_f = 32P_i[/itex] and [itex]V_f = V_i/8[/itex] the numerator in (1) is simply:

[tex]P_fV_f - P_iV_i = 32P_iV_i/8 - P_iV_i = 3P_iV_i[/tex]

for BC, [itex]P_f = P_i/32[/itex] and [itex]V_f = 8V_i[/itex] the numerator in (1) is simply:

[tex]P_fV_f - P_iV_i = 8P_iV_i/32 - P_iV_i = -3P_iV_i/4[/tex]

How did u know the gas was monoatomic?

Apply the adiabatic condition [itex]PV^\gamma = constant[/itex] to one of the adiabatic paths and solve for [itex]\gamma[/itex].

AM
 
Last edited:
  • #6
suspenc3
402
0
Thanks Andrew
 

Suggested for: Another Thermo Question

Replies
6
Views
90
Replies
15
Views
536
Replies
5
Views
410
Replies
9
Views
607
Replies
45
Views
2K
  • Last Post
Replies
11
Views
225
Top