Question

A cubic meter of air near saturation may contain 28 grams, or about 1.6 moles of water molecules at 30 °C, but

Answer 1

For the cubic meter of air near saturation that contains 28 grams (1.6 moles) of water molecules when T = 30°C and 8 grams (0.44 moles) of it when T = 8°C, we have:  

1. The pressure of air due to water vapor when the temperature is 30°C and 8°C, is 0.040 atm and 0.010 atm, respectively.            

2. Since the water vapor depends on the temperature, it is important to know the temperature of air so we can know how much air pressure is due to water vapor.                

     

1. The pressure of air near saturation can be calculated with the Ideal gas law equation:    

$PV = nRT$            

Where:      

P: is the pressure =?

V: is the volume = 1 m³ = 1000 L

n: is the number of moles

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

• Pressure when the temperature is 30°C

We have:              

T = 30 ° = 303 K

n: number of moles of water = 1.6 moles

Then the pressure is:

$P_{30} = \frac{nRT}{V} = \frac{1.6 \:moles*0.082 L*atm/(K*mol)*303 K}{1000 L} = 0.040 atm$

Hence, the pressure when the temperature is 30°C is 0.040 atm.

• Pressure when the temperature is 8°C

T = 8 °C = 281 K

n = 0.44 moles of water

The pressure is:

$P_{8} = \frac{nRT}{V} = \frac{0.44 \:moles*0.082 L*atm/(K*mol)*281 K}{1000 L} = 0.010 atm$

Therefore, the pressure when the temperature is 8°C is 0.010 atm.

2. Since the amount of mass or moles of water in the atmosphere depends on the temperature, having that the higher the temperature, the higher the number of water molecules present in the air, it is important to know the temperature so we can know how much of the air pressure is due to the water vapor and thus the value of humidity in the air.              

Find more here:          

brainly.com/question/1190311?referrer=searchResults      

I hope it helps you!