N__________ is the study of atoms.
2 answers:
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Explanation:
The given data is as follows.
Temperature of dry bulb of air =
Dew point = = (10 + 273) K = 283 K
At the dew point temperature, the first drop of water condenses out of air and then,
Partial pressure of water vapor = vapor pressure of water at a given temperature
Using Antoine's equation we get the following.
= 0.17079
= 1.18624 kPa
As total pressure = atmospheric pressure = 760 mm Hg
= 101..325 kPa
The absolute humidity of inlet air =
= 0.00735 kg / kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg / kg dry air.
Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.
0.07 kg - 0.00735 kg
= 0.06265 kg for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
= 6.265 kg
Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.
Answer:
The final temperature was 612 °C
Explanation:
Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:
When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:
In this case:
- V1= 5.76 L
- T1= 22 °C= 295 °K (Being 0°C=273°K)
- V2=17.28 L
- T2=?
Replacing:
Solving:
T2= 885 °K = 612 °C
<u><em>The final temperature was 612 °C</em></u>