Question

# In the $$P-V$$ diagram shown in figure $$ABC$$ is a semicircle. The work done in the process $$A \rightarrow B \rightarrow C$$ is

A
Zero
B
π2atmL
C
π2atmL
D
4 atm L

Solution

## The correct option is C $$\dfrac {\pi}{2}atm-L$$$$Given,\\\quad ABC\ is\ a\ semicircle\ of\ radius\ 1$$$$We\ know\ that,\ work\ done\ is\ area\ under\ the\ P-V\ graph$$$$For\ A\rightarrow B,\\ \Rightarrow \Delta V_{AB}=-ve\\ \Rightarrow \Delta W_{AB}=-ve\\$$$$For\ B\rightarrow C,\\ \Rightarrow \Delta V_{BC}=+ve\\ \Rightarrow \Delta W_{BC}=+ve$$$$Also,$$$$\qquad \left | W_{AB} \right |<\left | W_{BC} \right |$$$$Hence,\ total\ work\ done\ is\ positive.$$$$\Delta W_{ABC}=Area\ of\ semicircle$$$$\qquad\ \ \ \ \ \ =\dfrac{1}{2}\pi(1)^2$$$$\qquad\ \ \ \ \ \ =\dfrac{\pi}{2}\ atm-L$$$$Hence,\ option\ (B)\ is\ correct.$$Physics

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