Question

1-An open container with 2.0L of benzene (specific gravity of 0.8) is left under a laboratory hood.If the hood has an air volume of 12 m3 (i.e.,12000L), and equilibrium conditions are established almost immediately, what is the benzene concentration in the air? How much benzene remains as spilled liquid?If the hood is ventilated at the rate of 0.5m3/hr, how much of the benzene will remain as liquid after 1 hour? How long does it take for the concentrations under the hood to drop to 10% of the saturation level?

Answer 1

Hi
Since nothing is mentioned we can take the condition as the ambient conditions.

T=298 K and pressure =1atm

By use of Antonie cofficients we can find the value of the saturated vapor pressure of Benzene at this temperature.

lnPs = A- B/T+C
Here A,B, C are the constants and Ps is the saturated pressure in bar.

A=4.72
B=1660.65
C= -1.46

Keeping the values in the above equation we get
Ps =0.41 bar
External pressure = 1 atm =1.01 bar

So the mole fraction of the Benzene in the vapor =partial pressure of Benzene /Total pressure.

=0.41/1.01
=0.405

Volume of the room =12 m^3.
We can neglect the volume of the liquid.

We know that at 25 o C and 1 atm , one mole of gas occupies the total volume of 24.4 L =0.0244 m^3.

Moles of the total air mixture in room = volume / volume of one mole=12/0.0244=491.40 moles

Moles of benzene in the air = 491.40*0.405
=199 moles

Moles of the Benzene liquid available =Mass/Molecular mass = 2000*0.8/78 =20.51 moles

(Specific gravity is 0.8)

Since the required moles are less than that required for the saturated case and all the Benzne will go in the vapor phase.

When we are supplying air at a rate of 0.5 m^3 per hour.

Mole fraction of Benzene in the room =Moles of Benzene /Total moles
= 20.51/491.4
=0.041 mole benzne /mole of air.

Since in every hour we are replacing 0.5 m^3 of air with new air , concentration will become 11.5/12 times of initial value.

After t hours mole fraction of the Benzne =
0.041*(11.5/12)^t

Time required for dropping the value to the 10 % is t1 say.

0.041*(11.5/12)^t1=0.041/10
t1 =54 hours

Hope this helps

Thanks