Question

Work is done by a substance in reversible non-flow manner in accordance with

𝑉 = 100 𝑓𝑡3 where P is in psia. Evaluate the work done on or by the substance as the

𝑃

pressure increases from 10 psia to 100 psia

Answer 1

$Work\; done = \int Vdp \\ 1 \; ft^3 = 0.0283 \; m^3 \\ 1 \; psia = 6.895 \; kPa \\ 1 \;J = 1 \; N \cdot m = 0.73756 \;ft \cdot lbf \\ W = \int^{p_2}_{p_1}V dp \\ V = \frac{100}{p} \times 0.0283 \; m^3 \\ p_1 = 10 \times 6.895 = 68.95 \;kPa \\ p_2 = 100 \times 6.895 = 689.5 \; kPa \\ W = \int^{689.5}_{68.95} \frac{100}{p} \times 0.0283 dp \\ W = 2.83[ln(p)]^{689.5}_{68.95} \\ W = 6.516 \; kJ = 6.516 \times 10^3 \;J \\ W = 6.516 \times 10^3 \times 0.73756 \\ W = 4806.17 \; ft \cdot lbf$