By combining Boyle's and Charles' laws, an equation can be derived that gives the simultaneous effect of the changes of pressure and temperature on the volume of the gas. This is known as combined Ideal Gas Equation. A simple and direct method of deriving this equation is as follows:
According to Boyle's law, for a given mass of a gas at constant temperature
V a (1/P) (constant T) …….(i)
According to Charles' law,
V a T (constant P)…….(ii)
Combining (i) and (ii)
V a (T/P)
or
V = K(T/P)
or
(PV / T) =K= constant
Now, if V1 is the volume of a gas at temperature T1 and pressure P1. V2 is the volume of same amount of gas at temperature T2 and pressure P2, then
(P1V1/T1) = (P2V2 / T2) = constant
The above relationship is very useful for converting the volume of a gas from one set of conditions to another.
The numerical value of the constant of proportionality (K) depends upon the quantity of gas. The volume of a gas is directly proportional to the number of moles of gas at constant temperature and pressure (Avogadro's law). This means that 'K' is directly proportional to the number of moles, 'n', i.e.,
K a n
or k = nR
where 'R' is the universal gas constant of proportionality. The value of 'R' is same for all gases.
However, the numerical value of 'R' varies with the units in which pressure and volume are expressed. Therefore the Ideal gas Equation is derived as:
PV/T = nR or PV = nRT
For one mole (n = 1),
PV = RT (Ideal Gas Equation
The Ideal Gas Equation is also known as the equation of state for gases as it expresses the quantitative relation ship between the four variables that describe the state of the gas. The word 'ideal' is used here because in reality no gas obeys the above condition and the gases, which deviate from ideality are called as real gases.
Universal gas constant (R)
From the above ideal gas equation:
The universal gas constant is a measure of energy change (work done) per mole of the gas for one degree change in its temperature.
Numerical Value of R
The magnitude and unit of 'R' depends upon the units in which pressure, volume and temperature are expressed.
Conditions and problems
I)When pressure is expressed in atmosphere and volume in litres
Under standard condition of temperature and pressure i.e., when
P = 1 atm
T = 273.15 K
V = 22.414 L mol-1
This gives,
II) When pressure is expressed in atmosphere, and volume in mL
Here,
P = 1 atm
T = 273.15 K
V = 22414 mL mol-1
This gives,
III) R in energy units
When,
R = (PV/nT)
If 'R' is to be expressed in CGS units, the unit of pressure should be dyne cm-2 and volume in cm3 mol-1. Then,
Normal temperature T = 273.15 K
Normal pressure, P = (1 x 76 x 13.6 x 981) dyne cm-2
Molar volume under normal temperature and pressure
= 22414 cm3mol-1
Therefore,
= 8.314 x 107 erg deg-1 mol-1
Since, 4.182 x 107erg = 1calorie
Hence,
IV) In SI units
In the SI system of units, under NTP conditions,
P = 101325 N m-2
T = 273.15 K
V = 22.414x10-3 m3
Therefore,
= 8.314 J mol-1 K-1
The value of Gas constant R in different units
| 0.0821 | Litre atm K-1 mol-1 | P in atm, V in litre |
| 82.1 | atm cm3 K-1 mol-1 | P in atm, V in cm3 |
| 8.314x107 | Erg k-1 ,ol-1 | P in dynes cm-2, V in cm3 |
| 1.987 | Cal K-1 mol-1 | P in dynes cm-2, V in cm3 |
| 8.3144 | JK-1 mol-1 | P in pa or N m-2, V in |
Problems
5. Calculate the number of moles of hydrogen (H2) present in a 500 cm3 sample of hydrogen gas at a pressure of 760 mm of Hg and 27°C.
Solution
According to ideal gas equation, PV = nRT
P = 760mm of Hg = 760/760 = 1 atm, V = 500 cm3
T = 27 + 273 = 300 K, R=82.1 cm3 atm K-1 mol-1
n = PV/RT
n = 0.0203 mol = 2.03 x 10-2 mol.
6. About 200 cm3of a gas is confined in a vessel at 20°C and 740 mm Hg pressure. How much volume will it occupy at S.T.P.?
Solution
We are given
P1 = 740 mm Hg P2 = 760 mm Hg
T1 = 20 + 273 = 293K T2 = 273 K
V1 = 200 cm3 V2 = ?
According to gas equation,
Substituting the values, we get
= 181.4 cm3.
7. Calculate the volume occupied by 2 moles of an ideal gas at
2.5 x 105 Nm-2 pressure and 300 K temperature.
Solution
According to ideal gas equation,
PV = nRT or V = (nRT/P)
n = 2 mol, T = 300 K, P = 2.50 x 105 Nm-2
R = 8.314 Nm K-1 mol-1
= 19.95 x 10-3m3=19.95dm3.
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