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Deathcrush
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Homework Statement
A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure
P 1 to pressure P 2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas
constant of the gas is R, the entropy change of the gas ∆s during this process is
(a) ∆s =R ln(P 2 /P 1 ) (b) ∆s = R ln(P 2 /P 1 )- q/T (c) ∆s =R ln(P 1 /P 2 ) (d) ∆s =R ln(P 1 /P 2 )-q/T
(e) ∆s= 0
Homework Equations
PV=nRT (1)
u=q+w (2)
h=u+PV (3)
dw=-PdV (4)
ds=dq/T (5)
The Attempt at a Solution
first I considered h=0, since it is an isothermic process (I don't know if this is correct, this is one of my doubts)
I derived 2 and 3
du=dq+dw and du=-VdP-PdV
now, using 4 in this last derivative and comparing it with the first one, I found that dq=-VdP
and I used it in 5, also, I used 1 in V and integrated the whole thingS=R [tex]\int-dp/P)[/tex] from state 1 to state two
I integrated from state two to one, and changed the negative sign
this would be something like
S=R Ln(P1/P2)
after using the properties of the logarithms
My biggest doubt is that of the enthalpy, if it is not 0, the whole thing is wrong. I chequed the answers in the book and mine is right, but I don't know if the process is ok, would anybody help me on this please?
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