GASES
All gases resemble one another closely in their physically behavior.
Their volumes respond in almost exactly the same way to changes
in pressure, temperature, or amount of gas.
Measurements of gases:
Volume of gas:
A gas expands uniformly to fill any container in which it is placed.
This means that volume of a gas is the volume of its container.
Volumes of gases can be expressed in liters, cubic centimeters, or
cubic meters:
1L = 103 cm3 = 10-3 m3
Amount of gas:
Most commonly, the amount of matter in a gaseous sample is
expressed in terms of the number of moles (n). In some cases,
the mass (m) is given instead. These two quantities (n, m) are
related through the molar mass, M.
n= m / M
Temperature of gas:
Gas temperature is ordinarily measured using a thermometer
marked in degrees Celsius. However, in any calculation involving
the physical behavior of gases, temperatures must be expressed
on the kelvin scale. To convert between °C and k, use the
relation:
Tk = t °C + 273
Pressure of gas:
Pressure is defined as force per unit area. The pressure exerted by
a force of one Newton on 1 cm2 area.
A device commonly used to measure atmospheric pressure is the
mercury barometer.
The SI unit of pressure is the pascal (Pa), Atmospheric pressure is
about 105 Pa, or 100 kilopascals (kPa).
Another unit commonly used to express gas pressure is the
atmosphere (atm). This is the pressure exerted by a column of
mercury 760 mm high with the mercury at 0 °C.
1 atm = 760 mmHg
Problem: A balloon with a volume of 2.36 x 104m3 contains 4.68 x 106 g of
helium (MW= 4) at 18°C and 120 kPa. How can you express the following:
1. He volume in liters
2. He amount in moles
3. He temp in K
4. He pressure in atm & mmHg
Solution:
1m3 = 100 cm x 100 cm x 100 cm
=106 cm3
= 106 ml
= 103 L
The volume = 2.36 x 104 m3 x 103 = 2.36 x 107 L
The amount of He in moles = mass (m) / molar mass (M)
= 4.68 x 106 g / 4 g = 1.17 x 106 mol He
Tk = t °C + 273 = 18 + 273 = 291 k
1 atm = 101.3 kPa
The pressure (atm) = 120 kPa / 101.3 kPa = 1.185 atm
The pressure (mm Hg) = 1.185 atm x 760 mm Hg = 900.3 mm Hg
The ideal gas law:
All gases closely resemble each other in the dependence of volume on
amount, temperature and pressure.
Volume is directly proportional to amount
Vn
V = k1n (constant T, P)
where k1 is a constant
Volume is directly proportional to absolute temperature
V  T
V = k2T (constant n, P)
where k2 is a constant
The relationship was first suggested, in a different form by the balloonists
Charles and Gay-Lussac.
Volume is inversely proportional to pressure
V  1/P
V = k3 / P (constant n, T)
where k3 is a constant
The equation above is one form of Boyle's law.
The three equations relating the volume, pressure, temperature, and
amount of a gas can be combined into a single equation:
V  nT / P
V = constant ( n T / P)
Ordinarily, the constant is represented by the symbol (R).
Therefore, the ideal gas law is:
PV = n R T
Where: P is the pressure, V is the volume, n is the number of moles, T Kelvin
temperature, R is ideal gas constant, independent of P, V, n, T or the identity of the
gas.
Calculation of R constant:
R can be calculated from experimental values of P, V, n, and T.
Consider, for example, the situation that applies at 0 °C and 1 atm. This
conditions are often referred to as standard temperature and pressure for a
gas (STP). At STP, one mole of any gas occupies a volume of 22.4 L.
Solving the ideal gas law for R,
Substituting P = 1.00 atm, V = 22.4 L, n = 1 mol, T = 0 + 273 = 273 K
Problem1.
A 250-ml flask, open to the atmosphere, contains 0.0110 mole of air at 0 °C.
Upon heating, part of the air escapes; how much remains in the flask at 100
°C ?
Solution:
Initial state: P V = n1 R T1 so,
n1 T1 = PV / R ……………(1)
Final state: P V = n2 R T2 so,
n2 T2 = PV / R ……………(2)
From (1) and (2)
n1 T1 = n 2 T2
n1 = 0.0110 mole, T1 = 0 + 273 = 273 K,
T2 = 100 + 273 = 373 K
n1 T1 = n 2 T2
Thus, n2 = 0.0110 x 273 / 373 = 0.00805 mole
Problem 2.
If 2.5 gram of ozone (MW=102) introduced into an evacuated 500 ml container
at 10 °C, what pressure in atmospheres is developed?
Solution:
PV =nRT
V = 500.0 ml = 0.5 L
n = m / M = 2.5 g / 102.04 g C2F4H2 = 0.0245 mol
R = constant = 0.0821 L . atm. mol-1. K-1
T = 10 + 273 = 283 K
P=?
Thus,
P=nRT/V
= 0.0245 x 0.0821 x 283 / 0.5 = 1.14 atm
Problem 3.
Acetone is placed in a 3 L flask and vaporized by heating to 95 °C at 1.02 atm.
The vapor filling the flask at this temperature and pressure weighs 5.87 g.
Calculate the molar mass of acetone ?
Solution:
PV =nRT
,
n=m/M
PV =mRT/M
M=mRT/PV
= 5.87 x 0.0821 x (95 + 273) / 3.00 x 1.02
= 58 g / mole
Problem 4.
Taking the molar mass of dry air to be 29.0 g / mole, calculate the density of
air at 27 °C and 1 atm?
Solution:
PV =nRT
,
n=m/M
P V = m R T / M ……………………. (1)
d=m/V
m = d V ………………………………...(2)
PV =dVRT/M
PM =dRT
d = PM / R T = 1.00 x 29.0 / 0.821 x (27 + 273) = 1.18 g / L