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Andrew Mason
#5
Feb27-07, 04:31 AM
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Quote Quote by eaboujaoudeh View Post
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