Calculating virtual temperature

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The discussion revolves around calculating virtual temperature and water vapor in a classroom setting. For part a, the user successfully calculated the amount of water vapor present as 0.58 kg using the ideal gas law. However, in part b, the user struggles with calculating the virtual temperature, initially obtaining an incorrect value of 0.314 K instead of the expected 300.7 K. The confusion arises from incorrectly using the density of air as 1000 kg/m^3 instead of the correct value, which is approximately 1 kg/m^3. Clarification on the appropriate density for air is essential for accurate calculations.
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Homework Statement


Problem: On a summer day, your classroom warms and becomes muggy with a vapor pressure of 20hPa and a temperature of 25C.

a.) If the volume of the classroom is 40m^3, how much water is present in the room in vapor form?Assume density of liquid water is 1000 kg/m^3

b.) If the pressure in the room is 900hPa, what is the virtual temperature?

Homework Equations



Ideal gas laws: e=(m/v)RT ; P=ρ(Rd)Tv

The Attempt at a Solution



Finding the answer to part a.) was easy. I used the ideal gas law e=(m/v)RT and solved for m. The answer I got was .58kg, and that's correct. What I'm having a hard time with is part b.). I figured you could just use another form of the ideal gas law P=ρ(Rd)Tv, where Rd is the gas constant for try air, ρ is density, and Tv is the virtual temperature. So I set up my formula like this:

90,000 N/m^2 = (1000 kg/m^3)*(287 J/kg*K)*Tv

I converted the 900hPa into Pascals and then expressed Pascals as Newtons per square meter.

When I solve for Tv, I strangely get .314 K. I know this isn't right as the answer is supposed to be 300.7 K. So, what am I doing wrong here?
 
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Isn't the density of air only about one kilogram per cubic meter, rather than one tonne? That would explain why you're off by about a factor of 1000.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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