Combined gas law, ideal gas equation and universal gas constant

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

boyle s law

According to Charles' law,

Charles law

Combining (i) and (ii)

combination of boyle s and charles law



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

relation between pressure volume temperature

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 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:

ideal gas equation

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:

formula for universal gas constant

universal gas constant in terms of force and length

universal gas constant in terms of work

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,

value of R under STP


) When pressure is expressed in atmosphere, and volume in mL


P = 1 atm

T = 273.15 K

V = 22414 mL mol-1

This gives,

R value when pressure is  in atmosphere and volume in mL

III) R in energy units


If 'R' is to be expressed in CGS units, the unit of pressure should be dyne cm-2 and volume in cm3mol-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


value of R in energy units

= 8.314 x 107 erg deg-1 mol-1

Since, 4.182 x 107erg = 1calorie


value of R in calories

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


value of R in SI units

= 8.314 J mol-1 K-1

The value of Gas constant R in different units

value of gas constant


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.


According to ideal gas equation, PV = nRT

T = 27 + 273 = 300 K, R=82.1 cm3 atm K-1 mol-1

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.?


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.


According to ideal gas equation,

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.