Air contained in piston-cylinder arrangement, expands from 5 bar 300Cand 0.055m3 to a pressure of 2 bar according to law pv1.5=c. Calculate thework during the process
Question
Air contained in piston-cylinder arrangement, expands from 5 bar 300Cand 0.055m3 to a pressure of 2 bar according to law pv1.5=c. Calculate thework during the process
Solution
The work done during the process can be calculated using the formula for work done during an adiabatic process, which is given by:
W = (P2V2 - P1V1) / (n-1)
Where:
- P1 and V1 are the initial pressure and volume
- P2 and V2 are the final pressure and volume
- n is the polytropic index, which is 1.5 in this case
Given:
- P1 = 5 bar = 5 * 10^5 Pa
- V1 = 0.055 m^3
- P2 = 2 bar = 2 * 10^5 Pa
- n = 1.5
We need to find V2. We can use the equation of state for an adiabatic process, which is given by:
P1 * V1^n = P2 * V2^n
Solving for V2 gives:
V2 = (P1/P2)^(1/n) * V1 V2 = (5/2)^(1/1.5) * 0.055 = 0.089 m^3
Now we can substitute P1, V1, P2, V2 and n into the work done formula:
W = (210^5 * 0.089 - 510^5 * 0.055) / (1.5-1) W = -1.11 * 10^5 J
So, the work done during the process is -1.11 * 10^5 J. The negative sign indicates that the work is done by the system (expansion).
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