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Question 2.5: The properties of a closed system change as per the relation......

The properties of a closed system change as per the relation between pressure and volume as pV = 5 where p is in bar V is in m³. Calculate the work done when the pressure increases from 2 bar to 8 bar.

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pV = 5 ; p_1 = 2  bar ; p_2 = 8  bar

∴     V_1 = \frac{5}{2} \times 101.325  m^3 = 253.31  m^3 and V_2 = \frac{5}{8} \times 101.325  m^3 = 63.328  m^3

W_{12} = \int\limits_{1}^{2}{pdV} = \int\limits_{V_1}^{V_2}{\frac{5}{V} dV } = 5 \times 101.325 \log \frac{V_2}{V_1} = -702.2  kJ/kg

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