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0 votes
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in Thermodynamics by (15 points)
In the cyclic process shown on the V-P diagram, the numerical magnitude of the work done cannot be equal to i) \( \pi\left(\frac{P_{2}-P_{1}}{2}\right)^{2} \) ii \( \pi\left(\frac{V_{2}-V_{1}}{2}\right)^{2} \) iii) \( \frac{\pi}{4}\left( P _{2}- P _{1}\right)\left( V _{2}- V _{1}\right) \) iv) \( \pi\left( P _{2} V _{2}- P _{1} V _{1}\right) \) 1) (i), (ii), (iii) 2) (i), (ii), (iv) 3) only iv 4) All the four

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1 Answer

+1 vote
by (18.1k points)

Area under the curve = work done

Area of circle → πr2

Diameter = V2 - V1 = P2 - P1

Radius → \(\frac{V_2-V_1}2=\frac{P_2-P_1}2\)

(i) work done \(=\pi(\frac{P_2-P_1}2)^2\)

(ii) work done = \(\pi(\frac{V_2-V_1}2)^2\)

πr2 → π x \((\frac{V_2-V_1}2)\) \((\frac{P_2-P_1}2)\)

(iii) work done = π/4 (V2 - V1) (P2 - P1)

(iv) work done = π/4 (P2V2 - P1V2 - P2V1 + V1P1)

Option (1) is correct.

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