Find Work Done in Thermo Question | Monatomic Gas | Calculation Help

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The discussion focuses on calculating work done by a monatomic gas during various thermodynamic processes. Participants explore methods for finding the area under curves representing these processes, suggesting integration as a viable approach. The conversation highlights the use of specific equations for isentropic expansion and compression, emphasizing the relationship between pressure, volume, and the specific heat ratio (k). There is a clarification regarding the identification of the gas as monoatomic and the application of the adiabatic condition. The thread concludes with a reference to using the adiabatic condition to solve for the specific heat ratio in the context of the processes discussed.
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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:

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

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
 

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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.
 
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?
 
eaboujaoudeh said:
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) W = K\frac{V_f^{1-\gamma} - V_i^{1-\gamma}}{1-\gamma}

where K = PV^\gamma

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

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

P_fV_f - P_iV_i = 32P_iV_i/8 - P_iV_i = 3P_iV_i

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

P_fV_f - P_iV_i = 8P_iV_i/32 - P_iV_i = -3P_iV_i/4

How did u know the gas was monoatomic?

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

AM
 
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Thanks Andrew
 
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