One mole of an ideal gas is taken round the cycle A -> B -> C -> D -> A: A (2.00 bar, 10.0 L) -> B (2.00 bar, 30.0 L) isobaric; B -> C (5.00 bar, 30.0 L) isochoric; C -> D (5.00 bar, 10.0 L) isobaric; D -> A isochoric. Taking 1 L bar = 100 J, the net heat exchanged in one cycle is

6.0 kJ released to the surroundings

6.0 kJ absorbed from the surroundings

3.0 kJ released to the surroundings

zero, because Delta U = 0 for the cycle

6.0 kJ released to the surroundings

JEE Advanced levelSingle correct optionDifficulty 3 of 3td067

One mole of an ideal gas is taken round the cycle A $\to$ B $\to$ C $\to$ D $\to$ A: A ( $2.00 bar$ , $10.0 L$ ) $\to$ B ( $2.00 bar$ , $30.0 L$ ) isobaric; B $\to$ C ( $5.00 bar$ , $30.0 L$ ) isochoric; C $\to$ D ( $5.00 bar$ , $10.0 L$ ) isobaric; D $\to$ A isochoric. Taking $1 L bar = 100 J$ , the net heat exchanged in one cycle is

Work done by the gas on A

\to

B is

2 \times 20 = 40 L bar

; on C

\to

D it is

- 5 \times 20 = - 100 L bar

; the isochoric steps give zero. Net work by gas

= - 60 L bar = - 6.0 kJ

(the cycle runs anticlockwise on a

P

V

plot). For a cycle

Δ U = 0

, so

q_{n e t} = w_{b y g a s} = - 6.0 kJ

: 6.0 kJ of heat is released.