Question

Prove that in the PV graph of a thermodynamic cycle the ratio of slope of the adiabatic curve to the slope of isothermal curve is $\gamma $ .

Answer 1

Hint: The P-V equations for an isothermal process and adiabatic process is given by:
Isothermal process: ${\text{PV = constant}}$
Adiabatic process: ${\text{P}}{{\text{V}}^{\text{\gamma }}}{\text{ = constant}}$
And as we know slope of curve is given by:
$\tan \theta = \dfrac{{dy}}{{dx}}$
Hence, slope for P-V plot will be given by:
$\tan \theta = \dfrac{{dP}}{{dV}}$
Differentiate the above equations for both adiabatic and isothermal curves and find the value of $\dfrac{{dP}}{{dV}}$.
Then find the ratio of both the slopes

Complete step by step answer:
For an isothermal process, we know that temperature is constant throughout the process.
And from ideal gas law we know that:
$PV = nRT$
Where, n is the no. of moles of the gas.
R is the universal gas constant.
And T is temperature.
Since, here all the quantities n, R and T are constant.

Hence, PV=Constant. (This is the equation for the isothermal curve)
Differentiating the above equation, we get:
$
PV = c \\
\Rightarrow PdV + VdP = 0{\text{ [derivative of constant is zero]}} \\
\Rightarrow \dfrac{{dP}}{{dV}} = \dfrac{{ - P}}{V} \\
 $
Since, we know the slope of a curve is given by $\tan \theta = \dfrac{{dy}}{{dx}}$.
Hence, here slope of isothermal curve will be given by:
$\tan \theta = \dfrac{{dP}}{{dV}} = \dfrac{{ - P}}{V}$ …………….eq1
For an adiabatic process, we know that heat exchange is zero.
And it is defined by the equation:
${\text{P}}{{\text{V}}^{\text{\gamma }}}{\text{ = constant}}$
Differentiating, we get:
$
\Rightarrow (dP){V^\gamma } + Pd({V^\gamma }) = 0 \\
\Rightarrow (dP){V^\gamma } + \gamma P \cdot {V^{\gamma - 1}} \cdot dV = 0 \\
\Rightarrow \dfrac{{dP}}{{dV}} = - \gamma \dfrac{{P{V^{\gamma - 1}}}}{{{V^\gamma }}} = - \gamma \dfrac{P}{V} \\
 $
Hence, slope, $\tan \theta = \dfrac{{dP}}{{dV}} = - \gamma \dfrac{P}{V}$ ……………...eq2
Dividing equations 1 and 2, we get:
$\dfrac{{adiabatic}}{{isothermal}} = \dfrac{{ - \gamma \dfrac{P}{V}}}{{ - \dfrac{P}{V}}} = \gamma $
The slope of an adiabatic curve being $\gamma $ times the isothermal curve signifies that, for the same thermodynamic process the adiabatic curve is steeper than the isothermal curve.
Thus, from equations 1 and 2, we get the ratio of slopes of adiabatic and isothermal curves is $\gamma $

Note:
Take proper measures while differentiating the equations.
Here, the negative sign before the slopes for both the isothermal and adiabatic curves signifies that both the curves are decreasing (i.e. the curve is dipping downwards) and the angle made by the tangent with horizontal at any point on the curve is an obtuse angle ($\because \tan \theta $ for $\theta $$ > 90^\circ $ is negative).