Question

For the given pressure $\left(P\right)$ versus volume $\left(V\right)$ graph of an ideal gas for a cyclic process, the temperature of the gas will

• A. increase for the process $AB$
• B. increase for the process $BC$
• C. remain constant for the cyclic process
• D. decrease for the process $DA$

Answer 1

The correct option is A increase for the process $AB$

Using ideal gas equation, $PV=nRT$
For all the process, $n$ and $R$ remains constant.

For process AB: $P=\text{constant}$ and $V$ increases, therefore temperature$\left(T\right)$ increases.
For process BC: $P$ decreases and $V=\text{constant}$, therefore temperature$\left(T\right)$ decreases.
For process CD: $P=\text{constant}$ and $V$ decreases, therefore temperature$\left(T\right)$ decreases.
For process DA: $P$ increases and $V=\text{constant}$, therefore temperature$\left(T\right)$ increases.
Hence option $\left(a\right)$ is correct.

Alternative:
We know that $P-V$ curve for a constant temperature process $i.e$ isothermal condition is a hyperbola.


On visualizing these hyperbolic plots on the given $P-V$ curve,

Here, $T_1>T_2>T_3>T_4>T_5>T_6$
Thus, during the process $AB$ and $DA$, temperature increases and decreases for the process $BC$.

$\begin{array}{c}\text{Why this question?}\\ \text{Tips: Temperature relation is asked during various processes}\\ \text{in a}PV\text{diagram. Hence, it hints to visualize parallel hyperbolic curves}\\ \text{in the given diagram and temperature can be compared easily.}\end{array}$