Homework Statement
Solve DE: y' -2 y / (x + a) = -1 where a = constant
Homework Equations
y' + p(x) y = q(x) solving this DE with integrating factor.
The Attempt at a Solution
Use integrating factor (x+a) ^ -2 for above DE,
[ y'(x+a) ^ -2 ]' = - (x+a)^ -2 solving this DE we get
y =...
I think I managed to get the answer wrong in the last post.
I think the correct answer is (I hope this one is right):
p = (RF)/(kACv) - {(RF)/(kACv) - P0} (V0/V)^k
DE is:
p + V(dp/dV) = (R/Cv)(F/A) - (R/Cv)p,
then the solution to the DE is (I hope no mistakes):
p = (R/Cp)(F/A) [1 - (V0/V)^k] - P0(V0/V)^k (where k=Cp/Cv)
I believe that the DE corresponding to this problem is:
p + V(dp/dV) = (R/Cv)(F/A) - (R/Cv)p, (where (R/Cv) and (F/A) are constants )
solving this will get you p=p(v), however the solution is a bit more complicated than
(P-P0)V^k = Const.
My mistake was in the limits if...
Yes friction will produce irreversibilities. However solving the diff. eq.
dT/dV = -R/Cv (T/V) + F/(n Cv A)
and then obtaining the relation P=P(V) as I explained before will produce results which,
I think are inconsistent.
Your equation is correct,
dU = - {P - (F/A)}dV,
multiply out
dU = -PdV + (F/A)dV, (F/A is constant)
By the ideal gas law PV = nRT or P= nRT/V (n=number of moles of gas)
Replace "P" in equation for "dU"
dU = -(nRT/V)dV + (F/A)dV
But dU = (n Cv) dT (gas' internal...
An ideal gas is contained in a piston-cylinder arrangement. The walls of the cylinder and the piston itself are "adiabatic", i.e. no heat energy can be exchanged between the piston-cylinder and its surroundings. The piston experiences friction as it moves. All the heat generated by the...