Question

The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is: U = a + bPV where a = b = 3. The ratio of specific heats is (Answer up to two decimal places)

Answer 1

For an adiabatic process,
$dQ=dU+PdV=0\Rightarrow d[a+bPV]+PdV=0\Rightarrow bPdV+bVdP+PdV=0\Rightarrow (b+1)PdV+bVdP=0\Rightarrow (b+1)\frac{dV}{V}+b\frac{dP}{P}=0\Rightarrow (b+1)\log V+b\log P=constant;\log V^{b+1}+\log P^b=constantP{V}^{(b+1)/b}=constant\therefore \gamma =\frac{b+1}{b}=\frac{4}{3}=1.33$