Starting from the fundamental definitions for Cp and Cv, show that for all ideal gases,
Cp = Cv + R

I haven't learnt about entropy yet. This question should be possible just using the First Law and pV = RT.

I know Cv = du/dT|v and Cp = dh/dT|p


Also, can anyone explain to me what a quasi-eqilibrium process is?

consider a thermodynamic process in which a gas is heated.
case 1:constant volume:
dU=Q+dW
here dW = 0
hence dU=Q=Cp.dT... (1) (from definition)

case 2:at constant pressure, gas is heated so that the same change in internal energy takes place.
then dU = Q + dW
Q = Cv.dT
hence dU = Cv.dT + R.dT (dW = p.dV = R.dT)
substituting from (1), we get
Cp.dT = Cv.dT + R.dT
=>Cp = Cv + R,
which is the required result

A quasi-static process (meaning nearly static) or quasi equlibrium process is a process in which the system is in equilibrium throughout the process, i.e. the process is infinitly slow. The change in temperature and pressure of the system with its surroundings is infinitesimally small. It is just a hypothetical concept to apply the thermodynamical laws in an easy way.

There are mistakes from the answer. First law of thermodynamics is totally violated at the beginning and eq 1 should be Q = Cv dT, and some + and - sign misconception which at the end turn the answer around. It seems to me that the concept is totally wrong and it's just the math works. May check http://www.citycollegiate.com/molar_specific_heat2.htm for more info.

There are mistakes from the answer. First law of thermodynamics is totally violated at the beginning and eq 1 should be Q = Cv dT, and some + and - sign misconception which at the end turn the answer around. It seems to me that the concept is totally wrong and it's just the math works. May check http://www.citycollegiate.com/molar_specific_heat2.htm for more info.

to prove: cp-cv=r
cp and cv are molar specific heat capacities of an ideal gas at const. pressure and volume and r is universal gas constant.
To prove this relation we bigin with equation for 1 mole of gas
∆q=∆u+p∆v
if ∆q is absorbed at cnst.volume,∆v=0
Cv=(∆q/∆t)p=(∆u/∆t)v=(∆u/∆t)

Thread is old had question has already been replied in a clearer manner.

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