Question

One mole of a monoatomic ideal gas is taken along two cyclic processes $E\to F\to G\to E$ and $E\to F\to H\to E$ as shown on given $P-V$ diagram.
The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Which of the following is/are correct ?

• A. $W_{EFHE}>W_{EFGE}$
• B. $W_{EF}=0$
• C. $W_{EFHE}<W_{EFGE}$
• D. $W_{FH}<W_{FG}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $W_{EF}=0$
C $W_{EFHE}<W_{EFGE}$
D $W_{FH}<W_{FG}$

As work done is area under PV curve,

From graph, it can be deduced that,
$EF\to \text{isochoric}$
$W_{EF}=0$

$FH\to \text{adiabatic}$
$FG\to \text{isothermal}$
Area under $F-H<$ Area under $F-G$
$W_{FH}<W_{FG}$

Area of cyclic loop $E-F-H-E<$ Area of cyclic loop $E-F-G-H-E$
$W_{EFHE}<W_{EFGE}$