In This post you will get answer of Q.
An ideal monoatomic gas is taken round the cycle $ABCDA$ as shown in the P – V diagram (see figure). The work done during the cycle is
Solution:
Work done during the cycle
area enclosed by P-V graph
=area of $square ,A B C D $
$=A D times C D $
$=(2 V-V) times(2 P-P) $
$=PV$
Solution:
Work done during the cycle
area enclosed by P-V graph
=area of $square ,A B C D $
$=A D times C D $
$=(2 V-V) times(2 P-P) $
$=PV$