How Do You Calculate Moles of Water Vapor in Air Using Ideal Gas Law?

Click For Summary
SUMMARY

The calculation of moles of water vapor in air using the Ideal Gas Law involves determining the vapor pressure at a given temperature. In this case, at 30.0°C, the vapor pressure is 4200 Pa. By applying the Ideal Gas Law formula, n/v = p/RT, where R is the ideal gas constant (8.32 J/(mol·K)), the calculation yields approximately 1.67 mol/m³ of water vapor. This method effectively utilizes the relationship between temperature, vapor pressure, and the gas constant to derive the desired result.

PREREQUISITES
  • Understanding of the Ideal Gas Law
  • Knowledge of vapor pressure and relative humidity
  • Familiarity with temperature conversion to Kelvin
  • Basic proficiency in unit conversions (e.g., Pa to mol/m³)
NEXT STEPS
  • Study the Ideal Gas Law in detail, focusing on its applications in real-world scenarios
  • Learn about vapor pressure and how it varies with temperature
  • Explore the concept of relative humidity and its calculations
  • Investigate advanced gas law applications, including non-ideal gas behavior
USEFUL FOR

Students, educators, and professionals in chemistry, environmental science, and meteorology who are interested in understanding the calculations related to water vapor in the air and its implications for humidity and temperature.

owura143
Messages
12
Reaction score
0
The relative humidity is 55% on a day when the temperature is 30.0°C. Using the graph below, determine the number of moles of water vapor per cubic meter of air.


on the graph, Vapor pressure of water is on the y-axis and temperature on the x-axis 30C cordinates with 4200 Pa of water vapor.


This how i tried to sove it.

Since 30 corresponds to 4200 Pa

T in Kelvin =303
n/v = p/RT
=4200/(8.32 x 303)
= 1.67
 
Physics news on Phys.org
I need help with this same problem...

http://www.webassign.net/CJ/12_76.gif

That is the graph, if anyone can help.

EDIT... I really don't know where to start at all. I tried to do it (almost) the same way as owura, but I couldn't figure it out how to find it in "mol/m^3".
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
13
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K