Explanation:
The given data is as follows.
MW = 29 g/mol,
= 3.5 R
Formula to calculate minimum amount of work is as follows.
![W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1]](https://tex.z-dn.net/?f=W_%7Bs%7D%20%3D%20C_%7Bp%7DT_%7B1%7D%5B%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%5E%7B%5Cfrac%7BR%7D%7BC_%7Bp%7D%7D%7D%20-%201%5D)
= ![3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1]](https://tex.z-dn.net/?f=3.5%20%5Ctimes%208.314%20J%2Fk%20mol%20%5Ctimes%20306%20%5Ctimes%20%5B%28%5Cfrac%7B550%7D%7B125%7D%29%5E%7B%5Cfrac%7B1%7D%7B3.5%7D%7D%20-%201%5D)
= 4.692 kJ/mol
Therefore, total work done will be calculated as follows.
Total work done = 
Since, m =
. Therefore, putting these values into the above formula as follows.
Total work done = 
=
= 24268.96 kJ/min
It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.
= 402.86 kW
Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.
They are very stable (not reactive).
This is because their outer shell has a total of 8 valence electrons. All elements strive to get towards a full outer shell, but since these elements already have a full outer shell, they are fairly unreactive.
Weathering and erosion produce ever smaller rock particles which,
when mixed with dust and decayed organic matter over time ,
result in different types of soil .
[hope this helps]
Answer: When coal is burned, chemical potential energy is transformed into thermal energy, light energy, and sound energy. Only the thermal energy is used for electricity production. Light and sound energy dissipate into the environment, immediately reducing efficiency.
Hope this helps.