Answer:
[tex]p_g = 57.64 lb/in^2[/tex]
Explanation:
Given data:
temperature of water vapor is 200 degree F
pressure is 1 atm
\from the standard water temperature tables , at temperature 200 degree F
[tex]p_{sat} = 11.538 psia[/tex]
[tex]\omega_1 = \frac{0.622 \times 0.2 \times 11.538}{14.7 - (0.2\times 11.538)}[/tex]
[tex]\omega_1 = 0.115[/tex]
at temp 200 degree F and[tex] \omega _1 = 0.115[/tex]
[tex]\phi = 0.255[/tex]
so relative humidity is given as
[tex]\phi = \frac{p_v}{p_g}[/tex]
[tex]0.255 = \frac{14.7}{p_g}[/tex]
[tex]p_g = 57.64 lb/in^2[/tex]