Q135 / 160 · AP EAMCET 2009 · Physics

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 are $\begin{aligned} & Q_1=6000 \mathrm{~J}, \quad Q_2=-5500 \mathrm{~J} ; Q_3=-3000 \mathrm{~J} ; \\ & Q_4=3500 \mathrm{~J} \\ & W_1=2500 \mathrm{~J} ; \quad W_2=-1000 \mathrm{~J} ; \quad W_3=-1200 \mathrm{~J} ; \\ & W_4=x \mathrm{~J} . \end{aligned}$ 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

  • A
    $500 ; 7.5 \%$
  • B
    $700 ; 10.5 \%$
  • Correct answer
    C
    $1000 ; 21 \%$
  • D
    $1500 ; 15 \%$

Solution

From first law of thermodynamics $\begin{aligned} & \qquad Q=\Delta U+W \\ & \text { or } \quad \Delta U=Q-W \\ & \therefore \Delta U_1=Q_1-W_1=6000-2500=3500 \mathrm{~J} \\ & \Delta U_2=Q_2-W_2=-5500+1000=-4500 \mathrm{~J} \end{aligned}$ $\begin{aligned} & \Delta U_3=Q_3-W_3=-3000+1200=-1800 \mathrm{~J} \\ & \Delta U_4=Q_4-W_4=3500-x \end{aligned}$ For cyclic process $\Delta U=0$ $\begin{aligned} & \therefore \quad 3500-4500-1800+3500-x=0 \\ & \text { or } \\ & x=700 \mathrm{~J} \\ & \text { Efficiency, } \eta=\frac{\text { output }}{\text { input }} \times 100 \\ & =\frac{W_1+W_2+W_3+W_4}{Q_1+Q_4} \times 100 \\ & =\frac{(2500-1000-1200+700)}{6000+3500} \times 100 \\ & =\frac{1000}{9500} \times 100 \\ & \eta=10.5 \% \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

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