An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states $Q_{1}=6000 \mathrm{~J} ; Q_{2}=-5500 \mathrm{~J} ; \quad Q_{3}=-3000 \mathrm{~J}$ $Q_{4}=+3500 \mathrm{~J}$ $W_{1}=2500 \mathrm{~J} ; \quad…

Question

An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states $Q_{1}=6000 \mathrm{~J} ; Q_{2}=-5500 \mathrm{~J} ; \quad Q_{3}=-3000 \mathrm{~J}$
$Q_{4}=+3500 \mathrm{~J}$
$W_{1}=2500 \mathrm{~J} ; \quad W_{2}=-1000 \mathrm{~J} ; \quad W_{3}=-1200 \mathrm{~J}$
$W_{4}=x \mathrm{~J}$
The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$. The values
of $x$ and $\eta$ respectively are

Aicharya · Accepted Answer

$Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}$
$=6000-5500-3000+3500=+1000 \mathrm{~J}$
$W=W_{1}+W_{2}+W_{3}+W_{4}$
$=2500-1000-1200+x=+300+x$
In cyclic process, $\Delta U=0$ Now, $Q=\Delta U+W$
or $1000=0+(300+x)$
$\therefore x=700 \mathrm{~J}$
$\eta=\frac{W}{Q_{1}+Q_{4}}$
$=\quad \frac{1000}{6000+3500}=10.5 \%$ ^

An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and…

Solution