A reversible compression of 1 mol of an ideal gas in a piston/cylinder device results in a pressure increase from 1 bar to P2 an
d a temperature increase from 500 K to 1000 K. The path followed by the gas during compression is given by:
PV1.55 = constant
and the molar heat capacity of the gas is given by:
CP/R = 3.30 + 0.63x10-3 T (T in K)
Determine the final pressure and the heat transferred during the process.
PV1.55 = constant
and the molar heat capacity of the gas is given by:
CP/R = 3.30 + 0.63x10-3 T (T in K)
Determine the final pressure and the heat transferred during the process.
1 answer:
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Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if
then you have a thin walled cylinder
or using the diameter:
- if
then you have a thin walled cylinder
Answer:
Equilibrium Temperature is 382.71 K
Total entropy is 0.228 kJ/K
Solution:
As per the question:
Mass of the Aluminium block, M = 28 kg
Initial temperature of aluminium, = 273 + 140 = 413 K
Mass of Iron block, m = 36 kg
Temperature for iron block, = 273 + 60 = 333 K
At 400 k
Specific heat of Aluminium,
At room temperature
Specific heat of iron,
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
Thus
= 273 + 109.71 = 382.71 K
where
= Equilibrium temperature
Now,
To calculate the changer in entropy:
Now,
For Aluminium:
For Iron:
Thus