First slide

Heat engine

Question

An ideal gas is subjected to a cyclic process involving four thermodynamic states; the amounts of heat (Q) and work (W) involved in each of these states are
Q₁ = 6000 J, Q₂ = – 5500 J ; Q₃ = – 3000 J; Q₄ = 3500 J
W₁ = 2500 J; W₂ = – 1000 J; W₃ = – 1200 J; W₄ = xJ.
The ratio of the net work done by the gas to the total heat abosrbed by the gas is 'η'. The values of x and 'η' respetively are

Moderate

A

500 ; 7.5%
B

700 ; 10.5%
C

1000 ; 21%
D

1500 ; 15%

Solution

Incyclic process Δu = 0
Q_net = W_net
${Q_1} + {Q_2} + {Q_3} + {Q_4} = {w_1} + {w_2} + {w_3} + {w_4}$
6000–5500–3000+3500 = 2500–1000–1200+x
x = 700 J
$\therefore$ W_net = (2500 - 1000 - 1200 + 700)J
= 1000
Heat supplied Q₁ = (6000 + 3500)J
= 9500 J.
$\therefore \eta = \frac{{{W_{net}}}}{{{Q_1}}} = \frac{{1000}}{{9500}} \times 100\% = 10.5\%$

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