A jar contains a gas and a few drops of water at 'T' Kelvin temperature. The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1%. The vapour pressure of water at two temperatures are 30 and 25 mm of Hg respectively. Calculate the new pressure in the jar?
Correct answer: A. 817 mm of Hg
- A. 817 mm of Hg
- B. 792 mm of Hg
- C. 830 mm of Hg
- D. 724 mm of Hg
Explanation
This is the following solution: Pressure in the jar =830 mm of Hg.Pressure of moisture =30 mm of Hg.Pressure of dry gas =830−30=800 mm of Hg.Since the volume is constant and the temperature is changing, using the ideal gas equation, T1/P1 = T2/P2 ⇒ T1/800 = 0.99T1/P2 ⇒P2 =0.99×800=792 mm of Hg.Therefore, the total pressure =792+25=817 mm of Hg.
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About Evaporation, Boiling Point and Vapour Pressure
Evaporation is the escape of higher-energy molecules from a liquid surface, while vapour pressure is the pressure exerted by vapour in equilibrium with that liquid. The topic distinguishes evaporation from boiling and relates both to temperature, intermolecular forces, surface area, atmospheric pressure and the boiling point.
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