Dew Point worksheet

1. Relative humidity: 54.1%

1. Find the most moisture the air could hold at this temperature, as a pressure (es). The Magnus formula gives it

   $\text{es = 6.1094 × e^}\left(\frac{17.625\times 26}{243.04+26}\right)$$=33.55235\text{\hspace{0.17em}hPa}$

2. The moisture the air really holds equals that maximum at the dew point, so use the same formula there

   $\text{e = 6.1094 × e^}\left(\frac{17.625\times 16}{243.04+16}\right)$$=18.146243\text{\hspace{0.17em}hPa}$

3. Relative humidity compares the moisture the air holds with the most it could hold, as a percent

   $\frac{18.146243}{33.55235}\times 100$$=\mathrm{54.083374\%}$

4. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,814.624276}\times 18.01528}{8.314462618\times 299.15}$$=13.143304\text{\hspace{0.17em}g/m³}$

2. Relative humidity: 53.1%

1. Find the most moisture the air could hold at this temperature, as a pressure (es). The Magnus formula gives it

   $\text{es = 6.1094 × e^}\left(\frac{17.625\times 22}{243.04+22}\right)$$=26.385518\text{\hspace{0.17em}hPa}$

2. The moisture the air really holds equals that maximum at the dew point, so use the same formula there

   $\text{e = 6.1094 × e^}\left(\frac{17.625\times 12}{243.04+12}\right)$$=14.000741\text{\hspace{0.17em}hPa}$

3. Relative humidity compares the moisture the air holds with the most it could hold, as a percent

   $\frac{14.000741}{26.385518}\times 100$$=\mathrm{53.062218\%}$

4. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,400.07409}\times 18.01528}{8.314462618\times 295.15}$$=10.278153\text{\hspace{0.17em}g/m³}$

3. Dew point: 16.8 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{68}{100}\right)+\frac{17.625\times 23}{243.04+23}$$=1.13807455$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 1.13807455}{17.625-1.13807455}$$=16.77678714\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,906.593281}\times 18.01528}{8.314462618\times 296.15}$$=13.949324\text{\hspace{0.17em}g/m³}$

4. Dew point: 23.9 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{70}{100}\right)+\frac{17.625\times 30}{243.04+30}$$=1.5798545$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 1.5798545}{17.625-1.5798545}$$=23.93046783\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{2,965.655176}\times 18.01528}{8.314462618\times 303.15}$$=21.196781\text{\hspace{0.17em}g/m³}$

5. Dew point: 12.5 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{55}{100}\right)+\frac{17.625\times 22}{243.04+22}$$=0.86514972$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 0.86514972}{17.625-0.86514972}$$=12.54581539\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,451.203474}\times 18.01528}{8.314462618\times 295.15}$$=10.653501\text{\hspace{0.17em}g/m³}$

6. Dew point: 14.4 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{70}{100}\right)+\frac{17.625\times 20}{243.04+20}$$=0.98342542$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 0.98342542}{17.625-0.98342542}$$=14.3623257\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,633.408436}\times 18.01528}{8.314462618\times 293.15}$$=12.072903\text{\hspace{0.17em}g/m³}$

7. Dew point: 34.1 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{90}{100}\right)+\frac{17.625\times 36}{243.04+36}$$=2.16850703$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 2.16850703}{17.625-2.16850703}$$=34.09789981\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{5,342.80345}\times 18.01528}{8.314462618\times 309.15}$$=37.446116\text{\hspace{0.17em}g/m³}$

8. Dew point: 22.7 °C

1. Combine the humidity and temperature into one helper number, γ (gamma)

   $\text{ln}\left(\frac{65}{100}\right)+\frac{17.625\times 30}{243.04+30}$$=1.50574653$

2. Run the Magnus formula backward with γ to get the dew point (Td)

   $\frac{243.04\times 1.50574653}{17.625-1.50574653}$$=22.70307601\text{\hspace{0.17em}°C}$

3. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{2,753.822663}\times 18.01528}{8.314462618\times 303.15}$$=19.682725\text{\hspace{0.17em}g/m³}$

9. Relative humidity: 47%

1. Find the most moisture the air could hold at this temperature, as a pressure (es). The Magnus formula gives it

   $\text{es = 6.1094 × e^}\left(\frac{17.625\times 24}{243.04+24}\right)$$=29.780712\text{\hspace{0.17em}hPa}$

2. The moisture the air really holds equals that maximum at the dew point, so use the same formula there

   $\text{e = 6.1094 × e^}\left(\frac{17.625\times 12}{243.04+12}\right)$$=14.000741\text{\hspace{0.17em}hPa}$

3. Relative humidity compares the moisture the air holds with the most it could hold, as a percent

   $\frac{14.000741}{29.780712}\times 100$$=\mathrm{47.012782\%}$

4. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,400.07409}\times 18.01528}{8.314462618\times 297.15}$$=10.208974\text{\hspace{0.17em}g/m³}$

10. Relative humidity: 44%

1. Find the most moisture the air could hold at this temperature, as a pressure (es). The Magnus formula gives it

   $\text{es = 6.1094 × e^}\left(\frac{17.625\times 24}{243.04+24}\right)$$=29.780712\text{\hspace{0.17em}hPa}$

2. The moisture the air really holds equals that maximum at the dew point, so use the same formula there

   $\text{e = 6.1094 × e^}\left(\frac{17.625\times 11}{243.04+11}\right)$$=13.105026\text{\hspace{0.17em}hPa}$

3. Relative humidity compares the moisture the air holds with the most it could hold, as a percent

   $\frac{13.105026}{29.780712}\times 100$$=\mathrm{44.005079\%}$

4. Turn the vapor pressure into grams of water per cubic meter. M = 18.01528 g/mol is the weight of water per mole, and T is in kelvins

   $\frac{\mathrm{1,310.502555}\times 18.01528}{8.314462618\times 297.15}$$=9.555842\text{\hspace{0.17em}g/m³}$