Air that initially occupies .14 m^3 at a gauge pressure of 103.0 kPa is expanded

Question

Air that initially occupies .14 m^3 at a gauge pressure of 103.0 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. PLEASE SHOW WORK AND EXPLAIN STEPS FOR LIFESAVER RATING. DO NOT COPY PREVIOUS ANSWERS

Explanation / Answer

Work done BY the air is given by the integral W = ? p dV from Vi to Vf Step 1 isothermal expansion From ideal gas law follows p·V = n·R·T = constant so p·V = pi·Vi p = pi·Vi / V Hence: W = ? pi·Vi / V dV from Vi to Vf = pi·Vi · ln(Vf/Vi) For this problem pi = p_gauge + p_atmosphere = 103kPa + 101.3kPa = 204.3 kPa pf = 101.3kPa Vi = 140m³ Since pf·Vf = pi·Vi => Vf = Vi · (pi/pf) = 140m³ · (204.3kPa / 101.3kPa) = 282m³ So the work done in this step is W1 = 204300Pa · 140m³ · ln( 282m³ / 140m²) = 20.03×106J = 20.03MJ Step 2 isobaric compression W = ? p dV from Vi to Vf = p · ? dV from Vi to Vf = p · (Vf - Vi) here p = 101.3kPa Vf = 140m³ Vi = 282m³ So work done in this step is: W2 = 101300Pa · (140m³ - 282m³) = -14.38×106J = -14.38MJ The total work is W = W1 + W2 = 20.03 -14.38MJ = 5.65MJ

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