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

In the following figures,in which case is heat absorbed by the gas?

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

$A$ monoatomic gas is taken along path $AB$ as shown in the $V-P$ diagram. Calculate the change in internal energy of the system.

$A$ gas takes part in two processes in which it is heated from the same initial state $1$ to the same final temperature. The processes are shown on the $P-V$ diagram by the straight lines $1-2$ and $1-3$. Points $2$ and $3$ lie on the same isothermal curve. If $Q_1$ and $Q_2$ are the heat transferred along the two processes,then:

One mole of an ideal gas is taken from $A$ to $B$,from $B$ to $C$,and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure at $A$ is $P_{0}$ and volume is $V_{0}$. Then,the internal energy:

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