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 Homework Statement:

FIRST TYPE: REVERSIBLE PROCESS
SECOND TYPE: REVERSIBLE PROCESS
 Relevant Equations:
 costant temperature
FIRST TYPE: REVERSIBLE PROCESS At the temperature of 127 ° C, 1 L of CO2 is reversibly compressed from the pressure of 380 mmHg to that of 1 atm. Calculate the heat and labor exchanged assuming the gas is ideal. Q = L =  34.95 J
CONDUCT 380 mmHg = 0.5 atm L = P1 * V1 * ln (P1 / P2) = 0.5 * 1 * ln (0.5 / 1) =  0.34 L * atm = 0.34 * 101325/1000 = 35.11 J
SECOND TYPE: REVERSIBLE PROCESS Calculate the maximum work obtainable by making 50 L of oxygen expand isothermally (27 ° C) from 100 atm to 50 atm. Suppose the gas has ideal behavior. 349 * 10 ^ 3 J
PROCESS: L = P1 * V1 * ln (P1 / P2) = 100 * 50 * ln (100/50) = 3465 L * atm = 3465 * 101325/1000 = 351165 J
CONDUCT 380 mmHg = 0.5 atm L = P1 * V1 * ln (P1 / P2) = 0.5 * 1 * ln (0.5 / 1) =  0.34 L * atm = 0.34 * 101325/1000 = 35.11 J
SECOND TYPE: REVERSIBLE PROCESS Calculate the maximum work obtainable by making 50 L of oxygen expand isothermally (27 ° C) from 100 atm to 50 atm. Suppose the gas has ideal behavior. 349 * 10 ^ 3 J
PROCESS: L = P1 * V1 * ln (P1 / P2) = 100 * 50 * ln (100/50) = 3465 L * atm = 3465 * 101325/1000 = 351165 J