Question

Use the References to access important values if needed for this question. The air over an unknown liquid is saturated with the vapor of that liquid at 30 °C and a total pressure of 0.995 atm. Suppose that a sample of 6.30 L of the saturated air is collected and the vapor of the unknown liquid is removed from that sample by cooling and condensation. The pure air remaining occupies a volume of 4.14 L at -7 °C and 1.00 atm. Calculate the vapor pressure of the unknown liquid at 30 °C. atm Submit Answer 5 question attempts remaining

Answer 1

Given that pure air occupies volume of 4.14 L at -7 degrees and 1.00 atm pressure.

Now we are going to find volume of pure air at initial condition given :30 degrees and 0.995 atm

According to Gas Law; (PV)/T = constant (where P=Pressure, V=Volume and T=Temperature respectively)

so on applying gas law:

(1 * 4.14)/264 = (0.995 * V1)/303 (Temperatures are converted to kelvin by adding 273 to them and V1 is volume of pure air at 30 degrees)

On solving we get V1= 4.775L

Now we are doing is that calculating the pressure of pure air at 30 degrees and then subtracting it from total given pressure.

We will apply PV=nRT for pure air and find out the pressure.

Atomic weight of air= 29

Number of moles=(1.225 *4.775)/29 [Density of air is 1.225 Kg/m3]

So on applying PV=nRT

T=303

R=8.314

P=0.0554 atm

Therefore answer is : 0.995-0.0554=0.9396 atm.