Question

A cyclic process $ABCD$ , shown in $V - T$ diagram, is performed with a constant mass of an ideal gas. Show the same process on a $P - V$ diagram.

Accepted Answer

In the process $A \to B :$ For the line, $P - V$ we have the equation of the line $V = aT$ and both $V$ and $T$ increase.

Now, from the ideal gas equation we have $PV = nRT$

This implies, _{$P (aT) = nRT$}

$\Rightarrow P = (\frac{nR}{a}) =$ constant

So, the process is an isobaric process as the $V$ increases.

Then, in the $P - V$ diagram, the line $AB$ will be a straight line parallel to the $V$ axis.

Now, $B \to C :$ the process is isochoric as the volume is constant. For the line $BC, V$ is constant and $T$ is decreasing,

So, according to the gas equation, $PV = nRT .$

This implies that, $P = \frac{nRT}{constant},$ or $P \propto T ($ with $T$ decreasing $)$

Along the line $BC, P$ decreases as $V$ remains constant. So, in the $P - V$ diagram, the line $BC$ will be a straight line parallel to the $P$ axis.

And, in the process $C \to A$ the temperature is constant, thus it is an isothermal process.

So, from the ideal gas equation $PV = nRT =$ constant.

Therefore, in the $P - V$ curve, the line $CA$ will be a hyperbola with $P$ increasing.

Hence, the complete cycle on the $P - V$ diagram is shown in the figure.

B M Sharma Solutions for Chapter: Kinetic Theory of Gases, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinetic Theory of Gases, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Kinetic Theory of Gases, Exercise 1: ILLUSTRATIONS with Hints & Solutions

Learn and Prepare for any exam you want !