Heat And Thermodynamics Ques 181
- For the given cyclic process $C A B$ as shown for a gas, the work done is
(2019 Main, 12 Jan I)
(a) $5 J$
(b) $10 J$
(c) $1 J$
(d) $30 J$
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Answer:
Correct Answer: 181.(b)
Solution:
Formula:
- Key Idea
In a cyclic thermodynamic process work done $=$ area under $p-V$ diagram. Also in clockwise cycle, work done is positive.
In the given cyclic process, work done $=\oint p d V=$ area enclosed by the cycle $=\frac{1}{2} \times$ base $\times$ height of triangle $(C A B)$ made by cycle
$$ =\frac{1}{2}\left(V _2-V _1\right)\left(p _2-p _1\right) $$
From graph, given
$$ \begin{aligned} V _2 & =5 m^{3}, V _1=1 m^{3} \\ p _2 & =6 Pa, p _1=1 Pa \\ & =\frac{1}{2}(5-1)(6-1)=\frac{1}{2} \times 4 \times 5=10 J \end{aligned} $$
BETA
