Home
Class 11
CHEMISTRY
The pressure inside a jar containing a g...

The pressure inside a jar containing a gas and a few drops of water is 830 mm of Hg. Calculate the new pressure in the jar if temperature of the jar is reduced by 1%. The vapour pressure of water at two temperatures is 30 and 25 min of Hg respectively.

A

821.7 mm

B

817 mm

C

846.7 mm

D

767 mm

Text Solution

Verified by Experts

The correct Answer is:
B
`P_("gas") = P_("dry gas") + P_("moisture") * P_("dry gas ") = 830 = 30 = 800 mm`
Applying , `(P_1 )/(T_1) = (P_2)/(T_2) `
`P_1 = 800 mm, T_1 = T(say), P_2 = ? , T_2 = T-(T)/(100) = 0.99T`,
`P_2 = (800)/(T) xx 0.99T = 792mm`
`P_("gas")=P_("dry gas ") + P_("moisture")=792 + 25 = 817 mm`
Promotional Banner

Similar Questions

Explore conceptually related problems

A jar contains a gas and a few drops of water at TK 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 og Hg Calculate the new pressure in the jar .

A jar contains a gas and a few drops of water at absolute temperture T_(1). The pressure in the jar is 830mmof mercuryand 25mmof mercury. The temperature of the jar is reduced by 1%. The saturation vapour pressures of water at the two temperatures are 30mm of murcury and 25mm of mercury. Calculate the new pressue in the jar.

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be :

At 298 K , the vapour pressure of water is 23.75 mm Hg . Calculate the vapour pressure at the same temperature over 5% aqueous solution of urea. [CO(NH_(2))_(2)] .

The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to

The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to :