Answer to Question #143823 in Mechanical Engineering for Shayan Bhattacharjee

Question #143823
The properties of a closed system change following the relation between pressure and volume as pV= 3.0, where P is in bar V is in m^3. Calculate the work done when the pressure increases from 1.5 bar to 7.5 bar.
1
Expert's answer
2020-11-13T10:51:12-0500

The relation between pressure(P) and volume (V) of a closed system is given as

P×V=3=constantP\times V=3=constant

P=3VP=\frac{3}{V} ---------------------------------------(1)

P1×V1=P2×V2P_1\times V_1=P_2 \times V_2

when P=1.5 bar=1.5×105Pa\times 10^5 Pa => V=31.5×105=2×105\frac{3}{1.5\times 10^5}=2\times 10^{-5} m3m^3

when P=7.5 bar=1.5×105Pa\times 10^5 Pa => V=37.5×105=0.4×105m3\frac{3}{7.5\times 10^5}=0.4\times 10^{-5} m^3 m3m^3

The work done during the process is given by

W=V1V2PdV=2×1050.5×1053V]dV=3×ln(0.42)=3×1.6J=4.828JW=\int_{V_1 }^{V_2}PdV=\int_{2\times 10^{-5} }^{0.5\times 10^{-5}}\frac{3}{V}]dV=3\times ln(\frac{0.4}{2})=-3\times 1.6 J=-4.828J

Negative sign indicates work done on the system.

 


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Comments

Assignment Expert
23.12.20, 11:36

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Shayan Bhattacharjee
13.11.20, 18:22

Thank you so much assignmentexpert.com

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