what is the relative humidity if there is .5 moles of water vapor per 100 liters of air at 40 degrees celcius?
To calculate the relative humidity, we need to know the maximum amount of water vapor the air can hold at a given temperature and pressure. This is referred to as the "saturation vapor pressure."
First, let's find the saturation vapor pressure of water at 40 degrees Celsius. We can use a formula called the Antoine equation:
log(P) = A - (B / (T + C))
Where:
P is the saturation vapor pressure in mmHg
T is the temperature in degrees Celsius
A, B, and C are constants specific to the substance (in this case, water)
For water, the Antoine equation constants are:
A = 8.07131
B = 1730.63
C = 233.426
Plugging in the values:
log(P) = 8.07131 - (1730.63 / (40 + 233.426))
Solving for log(P):
log(P) = 4.1844
Taking the antilog of both sides, we can find P (saturation vapor pressure):
P = 10^4.1844
P ≈ 1437.7 mmHg
Now, we need to determine the actual vapor pressure of the air given the moles of water vapor and the volume of air.
First, let's convert 100 liters of air to the appropriate unit for pressure, which is mmHg. Assuming the air pressure is around 1 atm (760 mmHg), we can calculate the partial pressure of the water vapor:
Partial pressure of water vapor = (moles of water vapor / total moles of air) * total pressure
Partial pressure of water vapor = (0.5 moles / 100 liters) * 760 mmHg
Partial pressure of water vapor ≈ 3.8 mmHg
Finally, we can calculate the relative humidity:
Relative humidity = (actual vapor pressure / saturation vapor pressure) * 100
Relative humidity = (3.8 mmHg / 1437.7 mmHg) * 100
Relative humidity ≈ 0.26%
Therefore, the relative humidity in this case would be approximately 0.26%.