Ideal Gases
- This equation works (approximately) for all gases regardless of their Chemical Identity!
- The constant R is called the UNIVERSAL GAS CONSTANT, and is a fundamental conversion factor. When we first contact an advanced alien civilization, they will know a value of R but it will convert their temperature to their energy units, and will not be of great use to us. We already have enough different units for R (see below).
- The value and units of R depend on the units used in determining P, V, n and T.
- Temperature, T, must always be expressed on an absolute-temperature scale (K) (otherwise Charles' Law doesn't work)
- The quantity of gas, n, is normally expressed in moles. This is just a baker's dozen for expressing the number of molecules
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Numerical Example:
1.00 mol of gas at 1.00 atm of pressure at 0.00°C (273.15 K) occupies what Volume?
V = nRT/p
V = (1.00 mol)(0.0821 L atm/mol K)(273.15 K) / (1.00 atm)
Therefore: V = 22.4 L
Standard Temperature and Pressure (STP)
Another Numerical Example:
Hydrogen peroxide, H2O2, can decompose to form water and gaseous molecular oxygen. (Write a balanced reaction for this decomposition)
A sample of hydrogen peroxide is heated and the O2 gas produced is collected in a 750 ml flask. The pressure of the gas in the flask is 2.80 atmospheres and the temperature is recorded at 53.6 °C.
How many moles of O2 gas were produced?
How many moles of peroxide decomposed?
n = pV/RT
n = (2.80 atm * 0.750 L) / ((0.0821 L.atm/mol.K) * (53.6 + 273.15)K)
Note conversion to ABSOLUTE TEMPERATURE!
n = 0.0783 mol O2 were produced
n=0.157 mol peroxide decomposed
Boyle's law, Charles's law and Avogadro's law all represent special cases of the Ideal Gas law (equation)
pV = nRT
V = nRT / p
V= (nRT) * (1/p)
V= constant * (1/p)
(Boyle's law)
Check out this Boyle's Law Simulation If the quantity of gas and the pressure are held constant then:
pV = nRT
V = (nR/p) * T
V = constant * T
(Charles's law)
Check out this Charle's [sic] law Simulation If the temperature and pressure are held constant then:
pV = nRT
V = n * (RT/p)
V = constant * n
(Avogadro's law)
Suppose everything changes at once. One thing we are very sure of is that the gas constant, R, is in fact a constant. If we label the properties of the state of the gas initially by the subscript 1, then the state of the gas initially is defined by:
Numerical Example:
A 1.0 liter sample of air at room temperature (25 °C) and pressure (1 atm) is compressed to a volume of 3.3 ml at a pressure of 1000. atm. What is the temperature of the air sample?
If the sample did not change temperature, the increase in pressure by 1000 fold would decrease the volume 1000 fold to make the volume 1.0 ml.
But the actual volume is 3.3 times that. So the absolute temperature must have increased by a factor of 3.3.
Therefore, the temperature is 3.3 * (298) = 983K or 710°C
Mass Density has the units of mass per unit volume.
Number Density has the units of molecules (moles) per unit volume and is directly derived from the Ideal Gas Equation of State:
(n/V) = p/RT
Numerical Example:
What is the mass density of carbon tetrachloride vapor at 714 torr and 125°C?
The molar mass of CCl4 is 12.0 + (4*35.5) = 154 g/mol.
125°C is (273+125) = 398K.
714 torr is 714/760=0.9395 atm
The density of the gas is (154)(0.9395)/(0.0821)(398) = 4.43 g/l
John Dalton (1766-1844) - (gave us Dalton's atomic theory)
Let's denote Pt as the total pressure of the gas sample mixture and P1, P2, P3, etc. as the partial pressures of the component gases in the mixture. Clearly, Dalton's law would suggest:
Since the ideal gas equation works for any gas regardless of its identity, it should work for the gas sample as a whole, with the mole number simply the sum of the moles of each component in the mixture:
Numerical Example:
A gaseous mixture made from 10.0 g of oxygen and 15.00 g of methane is placed in a 10.0 L vessel at 25.0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel?
(15.00 g CH4)(1 mol/16.0 g) = 0.9375 mol CH4
V=10.0 l
T=(273.15+25.0)K=298.15K
pO2 = (0.3125)(0.08206)(298.15)/(10) = 0.765 atm
pCH4 = (0.9375)(0.08206)(298.15)/(10) = 2.29 atm
ptotal = 2.294 + 0.7646 = 3.06 atm
The ratio of the partial pressure of one component of a gas to the total pressure is:
a) A synthetic atmosphere is created by blending 2.00 mol percent CO2, 20.0 mol percent O2 and 78.0 mol percent N2. If the total pressure is 750 torr, calculate the partial pressure of the oxygen component.
Mole fraction of oxygen is (20/100) = 0.200
Therefore, partial pressure of oxygen = (0.200)(750 torr) = 150. torr
b) If 25.0 liters of this atmosphere, at 37.0°C, have to be produced, how many moles of O2 are needed?
PO2 = 150 torr (1 atm/760 torr) = 0.197 atm
V = 25.0 L
T = (273.15+37.0)K=310.15K
R = 0.0821 L.atm/mol.K
P V = n R T
n = (PV)/(RT) = (0.197 atm * 25.0 L)/(0.0821 L.atm/mol.K * 310K)
n = 0.194 mol
Example:
The synthesis of nitric acid involves the reaction of nitrogen dioxide gas with water:
(5.0 atm)(450 L) = n(0.0821 L.atm/mol.K)(295K) = 92.9 mol NO2
92.9 mol NO2 (2 HNO3 / 3 NO2) = 62 mol HNO3
Collecting Gases Over Water
When heated, potassium chlorate gives off oxygen through the reaction:
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A sample of KClO3 is partially decomposed, producing O2 gas that is collected over water. The volume of gas collected is 0.250 liter at 25 °C and 765 torr total pressure.
a) How many moles of O2 are collected?
PO2 = 765 - 23.76 = 741.24 torr
PO2 = 741.2 torr (1 atm/760 torr) = 0.9753 atm
PV = nRT
(0.9753 atm)(0.250 L) = n(0.08206 L atm/mol K)(273.15 + 25.0)K
n = 9.97 x 10-3 mol O2
6.64 x 10-3 mol KClO3 (122.6 g/mol) = 0.815 g KClO3
(1.00658 atm)(V) = (9.97 x 10-3 mol)(0.08206 L atm/mol K)(273.15 + 25)K
V = 0.242 L
P1V1 = P2V2
V2 = (0.250 L)*((741.2 torr)/(765 torr))
V2 = 0.242 L
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