Question

The relation between internal energy $U$, pressure $P$ and volume $V$ of a gas in an adiabatic process is given as
$U=a+bPV$
where $a$ and $b$ are constants. What is the value of adiabatic constant $\left(\gamma \right)$?

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

Answer 1

The correct option is B $\frac{b+1}{b}$

From first law of thermodynamics,
$dQ=dU+dW...\left(i\right)$
we can write small amount of work done as $dW=PdV$
For an adiabatic process heat supplied is zero i.e $dQ=0$
From Eq$\left(i\right)$,
$\Rightarrow dQ=dU+PdV$
$\Rightarrow 0=dU+PdV$
substituting value of $U$
$\Rightarrow d(a+bPV)+PdV=0...\left(ii\right)$
since $a\&b$ are constants, $da=0,db=0$
Applying chain rule for finding $d\left(bPV\right)$ and putting in Eq $\left(ii\right)$
$\Rightarrow bPdV+bVdP+PdV=0$
or $(b+1)PdV+bVdP=0$
Dividing both sides by $PV$,
$\Rightarrow (b+1)\frac{dV}{V}+b\frac{dP}{P}=0...\left(iii\right)$
Integrating Eq.$\left(iii\right)$ both sides,
$\Rightarrow (b+1)\log V+b\log P=\text{constant}$
$\Rightarrow \log V^{b+1}+\log P^b=\text{constant}$
Applying $\log m+\log n=\log \left(mn\right)$
$\Rightarrow \log \left(V^{b+1}{P}^b\right)=\text{constant}$
$\Rightarrow V^{b+1}{P}^b=\text{constant}$
Taking power $\frac{1}{b}$ both sides,
$\Rightarrow P{V}^{\frac{b+1}{b}}={\text{constant}}^{\frac{1}{b}}$
$\Rightarrow P{V}^{\frac{b+1}{b}}=\text{constant}...\left(iv\right)$
Comparing Eq. $\left(iv\right)$ with $P{V}^{\gamma }=\text{constant}$
$\therefore \gamma =\frac{b+1}{b}$