The figure shows the $P-V$ plot of an ideal gas taken through a cycle $ABCDA.$ The part $ABC$ is a semicircle and $CDA$ is half of an ellipse. Then,

  • A
    The process during the path $A \to B$ is isothermal
  • B
    Heat is absorbed by the gas during the path $B \to C \to D$
  • C
    Work done during the path $A \to B \to C$ is zero
  • D
    Positive work is done by the gas in the cycle $ABCDA$

Explore More

Similar Questions

The heat absorbed by a system in going through the given cyclic process is: (in $\,J$)

In the given figure, $1$ represents isobaric, $2$ represents isothermal, and $3$ represents adiabatic processes of an ideal gas. If $\Delta U_{1}, \Delta U_{2}, \Delta U_{3}$ are the changes in internal energy in these processes respectively, then:

$A$ system changes from the state $(P_1, V_1)$ to $(P_2, V_2)$ as shown in the figure. What is the work done by the system?

An ideal gas follows the path shown in the figure. The net work done in the whole cycle is

The work done by a gas as it is taken in a cyclic process (shown in the graph) is (in $PV$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo