- #1
lucaud
- 1
- 0
- 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