Question

a jar contains a gas and few drops of water at T k. the pressure in the jar is reduced by 1%.the water saturated vapour pressure of water at the two temperatures are 30mm and 25mm of mercury. then the new pressures in the jar will be?explain?​

Answer 1

Answer:

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,

P

1

T

1

=

P

2

T

2

⇒

800

T

1

=

P

2

0.99T

1

⇒P

2

=0.99×800=792 mm of Hg.

Therefore, the total pressure =792+25=817 mm of Hg.