Final answer:
The specific work done by the air on the surroundings when compressed from 123 kPa and 36°C to 882 kPa in a piston-cylinder device, treating air as an ideal gas, is approximately 434.30 kJ/kg.
Step-by-step explanation:
To calculate the specific work done by the air when compressed in a piston-cylinder device, we can use the formula for the work involved in an isentropic process for an ideal gas, which is given by:
W = n R T ln(P2/P1)
However, we need the answer in specific work (per unit mass), so we can modify the equation to:
W = (R T1 / (γ - 1)) * ln(P2/P1)
Where:
- W is the specific work
- R is the specific gas constant for air
- T1 is the initial temperature in Kelvin
- γ (gamma) is the ratio of specific heats (cp/cv)
- P1 is the initial pressure
- P2 is the final pressure
Given values are:
- R = 0.287 kJ/kg-K
- T1 = 36°C, which is 309.15 K (36 + 273.15)
- P1 = 123 kPa
- P2 = 882 kPa
- cp = 1.004 kJ/kg-K
- cv = 0.717 kJ/kg-K
First, we calculate γ = cp/cv = 1.004 / 0.717 ≈ 1.40. Now, we can substitute the values into the modified formula to find the specific work:
W = (0.287 kJ/kg-K * 309.15 K) / (1.40 - 1) * ln(882/123)
W ≈ (88.46955 kJ/kg) / 0.40 * ln(7.17)
W ≈ 220.58 kJ/kg * 1.968 = 434.30 kJ/kg
Therefore, the specific work done by the air on the surroundings when compressed from 123 kPa and 36°C to 882 kPa is approximately 434.30 kJ/kg.