Question

A gas mixture contains 20.0 mole% H₂O(v) and 80.0 mole% N₂.

The gas temperature and absolute pressure at the start of each of the three parts of this problem are 66°C and 700.0 mm Hg.

Ideal gas behavior may be assumed in every part of this problem.

a. If some of the gas mixture is put in a cylinder and slowly cooled at constant pressure, at what temperature would the first drop of liquid form? °C

b. If a 30.0 liter flask is filled with some of the gas mixture and sealed, and 30.0% of the water vapor in the flask is condensed, what volume would be occupied by the liquid water?  cm³

What would be the system temperature? °C

c. If the gas mixture is stored in a rigid-walled cylinder and a low-pressure weather front moves in and the barometric (atmosopheric) pressure drops from 760 mm Hg to 720.0 mm of Hg, what would be the new gauge pressure in the cylinder?  mm Hg

Answer 1

a) gas mixture containing 20 mol% water vapor and 80 mol% nitrogen.

Let us assume mole fraction of water vapor, y1 = 0.2

Mole fraction of nitrogen, y2 = 0.8

Gas temperature T = 66 C = 66 +273 = 339 K

Gas pressure P = 700 mmhg = 700/760 = 0.921 atm

Here we are gas mixture treated as ideal gas.

a) of gas mixture put into the cylinder and slowly cool down at constant pressure then first drop of liquid form at dew point temperature.

At dew point temperature pressure become saturation pressure and temperature become saturation temperature.

By using roults law for ideal gas,

Partial pressure of water vapor = mole fraction of water vapor * total pressure

P1= y1*P = 0.2*700 = 140 mmhg

This partial become saturation pressure at dew point and there is no longer cool down mixture it means no temperature drop below dew point if again cool down then condensation starts at constant pressure and constant temperature.

Psat= 140 mmhg

Tdew = Tsat

By steam table saturation pressure Psat = 140 mmhg

We get, Tsat = 58.57 C

hence dew point of the gas mixture Tdew = 58.57 C .

b) if total volume of sealed flask V = 30 L

total moles of gas mixture by ideal gas equation,

PV = nRT

n = PV/RT

P = 0.921 atm

V = 30L

R = 0.0821 L.atm/mol.k

T = 339 K

n = (0.921atm *30L)/(0.0821L.atm/mol.k)*339k

n = 0.9928 mol

Moles of gas mixture (O2 + N2) , n = 0.9928 mol

Moles of water vapor, nv = y1*n = 0.2*0.9928 = 0.1985 mol

Given 30 % of the water vapor in the flask is condensed,nL = 0.3*nv = 0.3*0.1985= 0.0595 mol

Liquid water moles, nL = 0.0595 mol

Mass of liquid water, mL = 18gm/mol * 0.0595 mol = 1.0722 gm

By steam table, at Psat = 140 mmhg

Specific Volume of liquid water  vL = 1.01635*10^-3 m3/kg

vL = 1.01635*10^-3 * 10^6cm3/10^3 gm = 1.01635 cm3/gm

Volume of liquid water VL = mL*vL = 1.0722gm*1.01635 cm3/gm = 1.0897 cm3

volume of liquid water VL = 1.0897 cm3

temperature of the system after condensing water T = 58.57 C because condensation at dew point and no temperature drop during phase change.

C) if the gas mixture stored in the rigid wall cylinder.

And here barometric pressure which is surrounding pressure of the rigid wall cylinder drops from 760 to 720 mmHg.

Gauge pressure Pg = Pabsolute - Patmospheric

Absolute pressure of gas mixture Pabsolute = 700 mmhg

Pg = 700 - 720 = -20 mmHg

gauge pressure Pg = -20 mmHg