Question
A jar contains a gas and a few drops of water at T K . The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1%. The saturated vapour pressures of water at the two temperatures are 30 and 25 mm of Hg. Calculate the new pressure in the jar.
Read 1 Solution.
At T K, Pgas = Pdry gas + Pmoisture = 830-30= 800 mm
Now at new temperature T1 = T - T/100 - 0.99T
Since V1=V2 ; P/T= const. or P1/T1 = P2/T2
Pdry gas =(800 x 0.99T) / T = 792 mm
Pgas = Pdry gas + Pmoisture = 792 + 25 = 817 mm of Hg
SWATI KAPOOR 10 year ago
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