According to the $PV$ diagram shown in the figure,a thermodynamic system is taken from state $A$ to state $B$ along the path $ACB$ and is brought back to state $A$ along the path $BDA$. The net work done during the complete cycle is given by the area of:

  • A
    $P_1ACB P_2P_1$
  • B
    $ACBB'A'A$
  • C
    $ACBDA$
  • D
    $ADBB'A'A$

Explore More

Similar Questions

The heat energy absorbed by a system in going through a cyclic process shown in the figure is:

Difficult
View Solution

In the following indicator diagram,the net amount of work done will be

An ideal monoatomic gas is taken through the thermodynamic states $A \to B \to C \to D$ via the paths shown in the figure. If $U_A, U_B, U_C$ and $U_D$ represent the internal energy of the gas in states $A, B, C$ and $D$ respectively,then which of the following is not true?

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then:

The $P-V$ graph of an ideal gas cycle is shown below. The adiabatic processes are described by:

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