Isobaric Process: del(H) = mC(v)dT + (P.dV)/J

[Amith2006]

Homework Statement

Consider 1 gram of an ideal gas undergoing isobaric process. Suppose del(H) be the amount of heat given to it. Then,
del(H) = dU + del(W)
del(H) = 1 x C(v)dT + (P.dV)/J
But del(H) = C(p)dT
P.dV = r.dT
C(p)dT = C(v)dT + (r.dT)/J
C(p) - C(v) = r/J

Homework Equations

The Attempt at a Solution

In the above derivation, when volume is changing, how can they take dU = mC(v)dT?
Here m = mass of gas,r = gas constant,J = Mechanical equivalent of heat

 

[Andrew Mason]

In the above derivation, when volume is changing, how can they take dU = mC(v)dT?
Here m = mass of gas,r = gas constant,J = Mechanical equivalent of heat

U is a function of temperature only. It does not depend on volume or pressure (although those will affect temperature, of course). dU is always = mC(v)dT

AM

 

[Amith2006]

That was a nice piece of information.Thanks.