Question

The ratio of slopes of adiabatic and isothermal curves is:
(A) $\gamma$
(B) $\dfrac{1}{\gamma }$
(C) ${\gamma ^2}$
(D) ${\gamma ^3}$

Answer 1

Hint
For an adiabatic curve, the equation is: $P{V^\gamma } = constant$ . Similarly, for an isothermal curve, the equation is: $PV = {\text{constant}}$ . Differentiating both the equations to obtain the slopes and further finding the ratio would give you the solution.
$\Rightarrow P{V^\gamma } = {\text{constant}}$ for an adiabatic process and $PV = {\text{constant}}$ for an isothermal process, where $P$ and $V$ are the pressure and volume $\gamma$ is the ratio of specific heats at constant pressure and at constant volume.

Complete step by step answer
An adiabatic curve shows the relation between pressure and volume when there is no transfer of heat between the system and the surrounding. The adiabatic curve is described by using $P{V^\gamma } = constant$ .
Similarly, an isothermal curve shows the relationship between the pressure and the volume when the temperature of the system is maintained constant. This is illustrated using $PV = {\text{constant}}$ as shown above.
Now, the slopes of the two curves can be found by differentiating both of the equations. So, on differentiating the first equation with the help of the chain rule, we get,
$\Rightarrow dP{V^\gamma } + P\gamma {V^{\gamma - 1}}dV = 0$
On rearranging the equation we get,
$\Rightarrow dP{V^\gamma } = - P\gamma {V^{ - 1}}dV$
Since the slope is given by $\dfrac{{dP}}{{dV}}$ and it can be obtained by again rearranging the above equation as:
$\Rightarrow \dfrac{{dP}}{{dV}} = \dfrac{{ - P\gamma {V^{\gamma - 1}}}}{{{V^\gamma }}} = - \gamma \dfrac{P}{V}$ .......(3)
Similarly, the slope of the isothermal curve can be obtained by differentiating the $PV = {\text{constant}}$ :
$\Rightarrow VdP + PdV = 0$
On rearranging to the get the slope $\dfrac{{dP}}{{dV}}$ , we write:
$\Rightarrow VdP = - PdV$
and therefore,
$\Rightarrow \dfrac{{dP}}{{dV}} = - \dfrac{P}{V}$ ......(4)
As we have obtained the slopes of both adiabatic and isothermal curves, now the only thing left is to find the ratio of both, and that can be obtained by taking the ratio of equations (3) and (4). Hence,
$\Rightarrow \dfrac{{{{\left( {dP/dV} \right)}_{adia}}}}{{{{\left( {dP/dV} \right)}_{iso}}}} = \dfrac{{ - \gamma \dfrac{P}{V}}}{{ - \dfrac{P}{V}}} = \gamma$
Thus, we obtained that the ratio of the slopes of adiabatic and isothermal curves is equal to $\gamma$ and the correct answer is option (A).

Note
We can observe from this solution that the slope of the adiabatic curve is $\gamma$ times more than that of an isothermal process. So, it can be said that the adiabatic process is more rapid than the isothermal process, and that is why the adiabatic curve is steeper than the isothermal curve.