Question

Air that initially occupies 0.90 m^3 at a gauge pressure of 110 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.)

Answer 1

Work Done in the isothermal expansion of an ideal gas is

Work Done in the isothermal expansion of an ideal gas is W_i = nRTln[P_1/P_2] �if we model air as an ideal gas, we can write: PV = nRT �W_i = P_1 V_1 ln[P_1/P_2] �therefore, �P_1 = P_gauge + P_atm = 110 kPa + 101. 35 kPa = 211. 3 kPa �P_2 = 101. 3 kPa �W_i = [211. 3 times 10^3] (0. 9) ln [211. 3 times 10^3/101. 3 times 10^3] = 139. 81 KJ �therefore, �P_1 V_1 = P_2 V_2 �therefore V_2 = (211. 3 times 10^3 times 0. 9)/(101. 3 times 10^3) = 1. 877 m^3 �and so the work done in the isobaric process will be: W_f = 101 times 10^3 [0. 9 - 1. 877] = -99 KJ �therefore the total Work Done will be: W = 139. 81 + (-99) = 40. 81 KJ

if we model air as an ideal gas, we can write: PV = nRT

therefore,

P₁ = P_gauge + P_atm = 110kPa + 101.3kPa = 211.3 kPa

P₂ = 101.3kPa

therefore, $W_i = [211.3\times 10^3](0.9)ln[\frac{211.3\times 10^3}{101.3\times 10^3}] = 139.81kJ$ .

P₁V₁ = P₂V₂

therefore V₂ = (211.3 x 10³ x 0.9)/(101.3 x 10³) = 1.877 m³

and so the work done in the isobaric process will be: W_f = 101 x 10³[0.9 - 1.877] = - 99kJ

therefore the total Work Done will be: W = 139.81 + ( - 99) = 40.81 kJ.