Heat and Thermodynamics 5 Question 53
54. An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are $Q _1=5960 J, Q _2=-5585 J, Q _3=-2980 J$ and $Q _4=3645 J$ respectively. The corresponding quantities of work involved are $W _1=2200 J, W _2=-825 J, W _3=-1100 J$ and $W _4$ respectively.
(1994, 6M)
(a) Find the value of $W _4$.
(b) What is the efficiency of the cycle?
Show Answer
Answer:
Correct Answer: 54. (a) $765 J$ (b) $10.82 %$
Solution:
- (a) In a cyclic process $\Delta U=0$
Therefore, $\quad Q _{\text {net }}=W _{\text {net }}$ or $Q _1+Q _2+Q _3+Q _4=W _1+W _2+W _3+W _4$
Hence, $W _4=\left(Q _1+Q _2+Q _3+Q _4\right)-\left(W _1+W _2+W _3\right)$
$$ ={(5960-5585-2980+3645) $$
$-(2200-825-1100)}$
or $\quad W _4=765 J$
(b) Efficiency,
$$ \begin{aligned} \eta & =\frac{\text { Total work done in the cycle }}{\text { Heat absorbed (positive heat) }} \times 100 \\ & =\left(\frac{W _1+W _2+W _3+W _4}{Q _1+Q _4}\right) \times 100 \\ & ={\frac{(2200-825-1100+765)}{5960+3645} } \times 100 \\ & =\frac{1040}{9605} \times 100 \\ \eta & =10.82 % \end{aligned} $$
NOTE
- From energy conservation
$W _{\text {net }}=Q _{\text {tre }}-Q _{\text {-ve }}$ (in a cycle)
$\therefore \quad \eta=\frac{W _{\text {net }}}{Q _{\text {tve }}} \times 100=\frac{\left(Q _{+v e}-Q _{-v e}\right)}{Q _{+v e}} \times 100=\left(1-\frac{Q _{-v e}}{Q _{+v e}}\right) \times 100$
In the above question
$$ \begin{aligned} & Q _{-v e}=\left|Q _2\right|+\left|Q _3\right|=(5585+2980) J=8565 J \\ & \text { and } \quad Q _{\text {tve }}=Q _1+Q _4=(5960+3645) J=9605 J \\ & \therefore \quad \eta=\left(1-\frac{8565}{9605}\right) \times 100 \\ & \eta=10.82 % \end{aligned} $$
