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 and b are positive constants. What is the value of $\gamma$?

• A. $\frac{a}{b}$
• B. $\frac{b+1}{b}$
• C. $\frac{a+1}{a}$
• D. $\frac{b}{a}$

Answer 1

Answer: (B)

$\frac{b+1}{b}$

In an adiabatic process no heat is exchange. i.e. $(dQ=0)$

$dQ=dU+PdV$

$0=dU+PdV$

$d(a+bPV)+PdV=0$ ... ($V=a+bPV$..(given))

$bPdV+bVdP+PdV=0$

$(b+1)PdV+bVdP=0$

$(b+1)\left(dV/V\right)+b\left(dP/P\right)=0$ ... (dividing both by PV)

Integrating we get,

$(b+1)\log V+b\left(\log P\right)=$ constant

$\log V^{(b+1)}+\log P^b=$ constant

$V^{(b+1)}\times P^b=$ constant

Multiply power by $\frac{1}{b}$

$P{V}^{\left(\right(b+1\left)/b\right)}=constant=P{V}^{\gamma }$

$\gamma =\frac{b+1}{b}$

Hence option b is correct.