Calculating virtual temperature

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SUMMARY

The discussion revolves around calculating virtual temperature using the ideal gas law in a classroom scenario with specific conditions. The vapor pressure is given as 20 hPa, and the temperature is 25°C. The volume of the classroom is 40 m³, leading to a calculated water vapor mass of 0.58 kg. The main issue arises in part b, where the user incorrectly applies the ideal gas law, resulting in an erroneous virtual temperature of 0.314 K instead of the expected 300.7 K due to a misunderstanding of air density.

PREREQUISITES
  • Understanding of the ideal gas law
  • Knowledge of vapor pressure and its units
  • Familiarity with temperature conversions between Celsius and Kelvin
  • Basic principles of density in gases and liquids
NEXT STEPS
  • Review the ideal gas law applications in thermodynamics
  • Learn about vapor pressure and its impact on humidity calculations
  • Study the concept of virtual temperature and its significance in meteorology
  • Explore density variations in gases at different temperatures and pressures
USEFUL FOR

This discussion is beneficial for students in physics or engineering, particularly those studying thermodynamics, meteorology, or environmental science, as well as educators seeking to clarify concepts related to gas laws and temperature calculations.

Timebomb3750
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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?
 
Physics news on Phys.org
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.
 

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