Question

One mole of a monoatomic ideal gas undergoes a cyclic process as shown in the figure (where, $V$ is the volume and $T$ is the temperature ). Which of the statement(s) given below is (are) true?

• A. Process $\text{I}$ is an isochoric process
• B. In process $\text{II}$, gas absorbs heat
• C. In process $\text{IV}$, gas release heat
• D. Process $\text{I}$ and $\text{II}$ are not isobaric

Answer 1

Answer: (B), (C), (D)

Process $\text{I}$ and $\text{II}$ are not isobaric


For option - $A$:
Isochoric process - process in which $\Delta V=0$.

But from graph, it is clear that $\Delta V\ne 0$, thus process $\text{I}$ is not isochoric.

For option -$B$:
From the graph we can deduce that , Process $\text{II}$ is isothermal expansion.

$\Delta U=0$ and $W>0$ ($\because$expansion)
So, from first law of thermodynamics
$\Delta Q>0$

Hence gas absorbs heat.

For option -$C$:
Process $\text{IV}$ is isothermal compression.

$\Delta U=0$ and $W<0$ ($\because$compression)
So, $\Delta Q<0$

Hence, gas releases heat.

For option -$D$:
we know that,

Isobaric process - Proces in which $\Delta p=0$

But from graph, $\Delta p\ne 0$ as from ideal gas equation,
$pV=nRT$
For constant $p$ (i.e. $\Delta p=0)$
and we get,

$V\propto T$ [straight line]

Process $\text{I}$ and $\text{III}$ are not straight lines. Thus, these processes are not isobaric.

Hence, options (b) , (c) and (d) are the correct alternatives.


$\begin{array}{c}\text{Why this question}?\\ \text{Concept involved - (1) Different processes in thermodynamics}\\ \text{(2) First law of thermodynamics}\\ \\ \text{This problem required a blend concept}\\ \text{of graph and thermodynamics}\\ \text{Tip: - For such kind of problem, we can}\\ \text{check the options.}\end{array}$