Recent content by MACH2

Thank you for the reply.

(x+a)^-2 = (x+a) elevated to the -2 power (I am not used to this sites equation editor) Thanks for your reply,

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 =...

Very good Philip! Thanks for your interest.

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...

Indeed, getting rid of T an an earlier stage makes things more simple.

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...