Air that initially occupies 0.144 m3 at a gauge pressure of 96.3 kPa is expanded

Question

Air that initially occupies 0.144 m3 at a gauge pressure of 96.3 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)
1 J

Explanation / Answer

a similar question is solved below, but with different values. hope this helps you. Air that initially occupies 0.138 m3 at a gauge pressure of 107.1 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.) Work done by the air is given by the integral W = ? p dV from initial final state In first step we expand at constant temperature. i.e. p·V = n·R·T = constant hence: p = p1·V1/V and W1 = ? p1·V1/V dV from V1 to V2 = p1·V1·ln( V2/V1) because p1·V1 = p2·V2 V2/V1 = p1/p2 W1 = p1·V1·ln(p1/p2) Add standard atmospheric pressure to initial gauge pressure: p1 = 107.1kPa + 101.3kPa = 208.4kPa W1 = 208400Pa · 0.138m³ · ln(208.4/101.3) = 20746J In second step we operate at constant pressure W2 = ? p dV from V2 to V1 = p2 ? dV from V2 to V1 = p2 · (V1 - V2 ) with p1·V1 = p2·V2 => V2 = (p1/p2·)·V1 W2 = p2·V1· (1 - (p1/p2·)) = V1· (p2 - p1) = 0.0138m³· (101300Pa - 208400Pa) = -14780J So the net work done by the gas is W_tot = W1 + W2 = 20746J - 14780J = 5966J

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