- #1
binbagsss
- 1,326
- 12
Homework Statement
Calculate the change in internal energy of an ideal gas the expands reverisbly and adiabatically from ##v_i## to ##v_f##
Homework Equations
##pV^{\gamma} = constant ##*, where ##\gamma=\frac{c_p}{c_v}##
for a reversible adiabatic.
##E=\alpha p V##, ##\alpha## a constant.
The Attempt at a Solution
The solution does
##E=\alpha pV##
##\implies## ##E_f-E_i=\alpha (p_fV_f-p_iV_i) = W##
(And then continues to eliminate a variable using * which I am fine with etc)
MY QUESTION IS
(by the first law) ##dE=dW=-pdV##
But I don't understand how we can simply do ##p_fV_f-p_iV_i ## because I thought that, in general, ##p=p(V)## and so we need to integrate over ##p## instead of just looking at the initial and final state.(unless a process is isochoric - ##p(vf-vi)=W##).
Or in general would this be the case but due to * we do not need to integrate ? (if so how can this be seen)
Many thanks in advance.