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11.Thermodynamics
easy
An ideal gas is taken around $ABCA$ as shown in the above $P-V$ diagram. The work done during a cycle is
A
$2PV$
B
$PV$
C
$1/2PV$
D
Zero
Solution
(a) Work done = Area enclosed by triangle $ABC$
$ = \frac{1}{2}AC \times BC = \frac{1}{2} \times (3V – V) \times (3P – P) = 2\,PV$
Standard 11
Physics
