Work is done by a substance in a reversible nonflow manner in accordance with v = (700/p) m3, where p is in kPaa. Evaluate the work done on or by the substance as the pressure increases from 70 kPaa to 700 kPaa.
W = kJ
Answer in 2 Decimals
Solution;
"V=(\\frac{700}{P})m^3"
Differentiate;
"dV=\\frac{-700}{P^2}dP"
Work is calculated as;
"W=\\int PdV"
"W=\\int P(\\frac{-700}{P^2})dP"
"W=-700\\int\\frac1PdP"
"W=-700[ln(P)]_{70}^{700}"
"W=-700[ln(700)-ln(70)]"
"W=-1611.809565kJ"
Answer;
W=-1611.81kJ
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