Show that in a reversible adiabatic process, for 1 mole of ideal gas:

$$P_1V_1^ \gamma=P_2V_2^ \gamma$$

Where$$\gamma =\frac{C_P}{C_V}$$

Question

Show that in a reversible adiabatic process, for 1 mole of ideal gas:

$$P_1V_1^ \gamma=P_2V_2^ \gamma$$

Where$$\gamma =\frac{C_P}{C_V}$$

anonymous · Answer

OK, it's adiabatic, so dQ = 0, so dE = -dW.  Set dW = p dV and dE (for an ideal gas) equal to Cv dT.  Now use the ideal gas equation of state (pV = nRT) to express p in terms of V.  Separate variables, so you have T and dT on one side, and V and dV on the other. Integrate both sides.  You'll then need R = Cp - Cv for an ideal gas.