One mole of liquid water is in a closed vessel whose temperature is 273 k, whose initial internal pressure is 1 atm (the space above the liquid is filled with air), and whose interior volume is 50 times the volume of the water. calculate the volume of the vessel. now the vessel is heated to the boiling point of water. assuming that the gases are ideal, calculate the pressure in the vessel.

1) volume of the vessel = 22.86 L

2) the pressure in the vessel  = 1.37 atm

Explanation:

to Calculate the volume of the vessel we are going to use this formula :

PV = nRT

when :

P is a pressure which is the force per unit area extended by the gas on the vessel = 1 atm

V is a volume which is the amount of space occupied by air or gas= ??

n is the number of moles = 1 mole

R is the ideal gas constant = 0.08206 L atm mol-1K-1

T is temperature in kelvin =273 K

by substitution:

1 atm * V = 1* 0.08206 * 273

V of air = 22.4 L

to assume V of water we can make this equation:

volume of the vessel = V of water + V of air

by Assuming V of water = X &  volume of the vessel = 50 X

so ,

50 X = X + 22.4 L

X = 22.4 L/49 = 0.457 L

so,volume of the vessel = 0.457 *50 or ( 0.457+22.4) = 22.86 L

2) to calculate the pressure in the vessel at constant volume:

P1/T1 = P2/T2

at constant volume the relation between pressure and temperature is directly proportional

when  the boiling point of water is equal = 100 °C = 100 + 273 = 373 K

1 atm /273 K = P2 / 373K

So P2 = 373/273*1

          = 1.37 atm

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the correct answer is:

a. formation of precipitation