Ideal Gas Equation
Table of Contents
- Introduction to Ideal Gas Equation
- Derivation of Ideal Gas Equation
- Alternative Derivation of Ideal Gas Equation.
- Ideal Gas Equation in terms of Density
- Nature and Values of the Gas Constant R
- Numerical Value of the Gas Constant (R)
- Dalton’s Law of Partial Pressures
- Applications of Dalton’s Law of Partial Pressures
Boyle’s Law gives the effect of pressure on the volume of a gas at constant temperature, while the effect of temperature on the volume of a gas at constant pressure is stated by Charle’s Law. However, one can derive an equation which gives the simultaneous effect of pressure and temperature on the volume of a gas.
The equation which gives the simultaneous effect of pressure and temperature on the volume of gas is known as Ideal Gas Equation or Equation of State for an Ideal Gas.
The gas equation may be derived from Boyle’s and Charles’ Law.
Let the volume of a certain mass of a gas change from V1 to V2 when the pressure is changed from P1 to P2 and temperature from T1 to T2 in two steps as shown in figure below:
Step 1: First suppose that the volume of a given mass of a gas changes from V1, to v when the pressure is changed from P1 to P2 at constant temperature T1.
Then according to Boyle’s Law,
Step 2: Now suppose that the volume v changes to V2 when the temperature is changed from T1 to T2 at constant pressure P2.
Substituting the value of v from Equation no. 1 in Equation no. 2, we have
This is the most convenient form of the ideal gas equation in calculations when out of 6 variables, one of them is unknown. This equation is also known as Combined Gas Law.
Another form of the ideal gas equation can be obtained as follows:
From Equation No. 2 (P × V)/T = Constant = K
The value of the constant K depends only upon the amount of the gas taken. If n is the number of moles of the gas taken, then
K ∝ n or K = nR
where R is a constant of proportionality which depends upon the amount of the gas taken and is found to be independent of the nature of the gas. The value of R is the same for 1 mole of any gas. Hence, R is called ‘Universal Gas Constant’.
Substituting the value of K in Equation No.3, we get
PV/T = nR or PV = nRT
This is the most common form of the ideal gas equation.
The ideal gas equation can be derived directly by combining Boyle’s Law, Charles’ Law and Avogadro’s law as follows:
According to Boyle’s law,
V ∝ 1/P at constant temperature Equation No. 1
According to Charles’ Law,
V ∝ T at constant pressure Equation No. 2
According to Avogadro’s law,
V ∝ n at constant temperature and pressure Equation No. 3
where number of moles of the gas is denoted by n. (This is because according to Avogadro’s law, when temperature and pressure are same, equal volumes of different gases will contain equal number of molecules or we can say that volume is directly proportional to the number of molecules which in turn is directly proportional to the number of moles of the gas.)
Combining Equation No.1,2 and 3 we get
Where, as before, R is molar gas constant. For 1 mole of the gas, the ideal gas equation becomes: PV = RT
A gas that obeys ideal gas equation exactly is called an ideal gas. An ideal gas equation defines the state of the gas completely when all the variables have been specified and because of this reason it is also called Equation of State.
This form of the ideal gas equation may be derived as follows:
If m represents mass of the gas in grams and M represents molecular mass of the gas, then
This expression can be used for the calculation of molecular mass from a known Value of Density of the Gas at a given temperature and pressure. Alternatively, the above equation can be used for the calculation of any one quantity from known values of the other quantities involved.
From Equation No.1 It may be concluded that under similar conditions of temperature and pressure, of different gases are directly proportional to their molar masses.
From the gas equation, PV = n RT, we have
(i) At NTP conditions, for 1 mole of the gas
P = 1 atmosphere, V = 22.4 liters, T = 273 K, n = 1 mole
= 0.0821 liter atmosphere degree-1 mol-1
If V = 22400 cm3, R = 82.1 cm3 atmosphere degree-1 mol-1
or Taking P = 1 bar so that V = 22.7 dm3,
= 0.083 bar dm3 K-1 mol-1
(ii) In the C.G.S. units, for 1 mole of the gas at NTP
P = 76 cm = 76 x 13.6 x 981 dynes/sq. cm
V = 22400ml, T = 273K, n = 1 mole
(iii) For Expressing R in S.I units, put 107 = 1 joule
= 8.314 joules degrees-1 mol-1 = 8.314 Jk-1 mol-1
Directly Taking P = 101.325 kPa, V = 22.4 dm3, T = 273K
We get, R = 8.314 kPa dm3 k-1 mol-1
(iv) To express in terms of calories, we know that
4.184 joules = 1 calorie
= 1.987 calorie degree-1 mol-1
Therefore after concluding from above we get the following table:
Units of P
Units of V
Values of R
Liters or dm3
|0.0821 liter atm degree-1 mol-1|
|82.1 cm3 atm degree-1 mol-1|
|8.314 × 107 ergs k-1 mol-1|
N m-2 or Pa
|8.314 Jk-1 mol-1|
|8.314 kPa dm3 K-1 mol-1|
|0.08314 bar dm3 k-1 mol-1|
Most of the gas laws usually discusses about the pressure, volume and temperature of a single gas. But pressures exerted by the mixtures i.e. if two or more chemically non-reacting gases are present in the same vessel, this property was discovered by John Dalton in 1807. He named the result known after himself known as Dalton’s Law of partial pressures:
If in an enclosed vessel, two or more gases (which do not react chemically) are present then the total pressure exerted by the gaseous mixture is equal to the sum of all the partial pressures that each gas would exert when present alone in the same vessel at the same temperature.
The pressure which that gas would exert when present alone in the same space at the same temperature is called Partial Pressure of a particular gas in a gaseous mixture enclosed in a given space. Let p1, p2, p3...pn be the partial pressures of n gases enclosed in a given vessel and P be the total pressure of the gaseous mixture. Then by Dalton’s Law of Partial Pressures, we have
P = p1+ p2+ p3+…...pn
Suppose we enclose ‘a’ moles of oxygen in a vessel at a given temperature and let its pressure be 159 mm of Hg. Now enclose ‘b’ moles of nitrogen in the same vessel at the same temperature. Let its pressure be 593 mm of Hg. But, the total pressure exerted by the gaseous mixture would be 752 mm Hg if at the same temperature and in same vessel, the same quantities of the two gases are enclosed (as shown in Figure no. 2) in accordance with the Dalton’s Law of Partial Pressures.
In the determination of pressure of a dry gas
Because saturated water vapours exert their own pressure, the gas which is collected over water it is moist. The pressure due to water vapours is called Aqueous Tension.
When the volume of the gas is measured after careful levelling (so that the level inside the measuring tube and outside in the Jar are same), the total pressure exerted by the gas and the water vapours is equal to the atmospheric pressure (fig. no. 3b)
Fig. No. 3 (a) Collecting a gas over water (b) Shifting the graduated tube by closing its lower end with a thumb into a jar and raising or lowering it so that the level of water inside the tube is same as that outside in the jar
If P and P’ are the pressure of the moist gas and the dry gas respectively at t C and p is the aqueous tension at that temperature, then by Dalton’s Law of Partial Pressures
P = P’ + p or P’ = P - p,
Pdry gas = Pmoist gas – Aqueous tension (at t C)
In the calculation of partial pressures
In a mixture of non-reacting gases A, B, C etc. If gas is considered to be an ideal gas, then applying PV = nRT, we have
By Dalton’s law of partial pressures,
Total pressure, P = pA + pB + pC +…...= (na + nb + nc +… )
Or pA = xA P
Similarly pB = xB P and so on. Thus, in a mixture of non-reacting gases, A, B, C etc.
Partial Pressure of A = Mole fraction of A x Total pressure
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