Question

Derive the expression of pressure of an ideal gas from kinetic theory of gas

Answer 1

From kinetic theory of gases, we have

P=13MVˉc2

or,PV=13Mˉc2

Since ˉC2∝T, if the temperature of the gas is kept constant, for given mass of a gas (i.e. M = constant),

PV=constant

i.e.p∝1V

It means pressure of the given mass of a gas is inversely proportional to it's volume at constant temperatures which is Boyle's law.

Charle's Law from Kinetic theory of gases

From kinetic theory of gases, we have

P=13MVˉc2

or,V=13MPˉc2

Since ˉC2∝T, if the pressure of the gas is kept constant, for given mass of a gas, we have

v∝T

It means volume of the given mass of a gas is directly proportional to it's absolute temperature at a constant pressure which is Charle's law.

Law of pressure from Kinetic theory of gases

here,

P=13MVˉc2

Since ˉC2∝T and at constant volume of a gas, the above equation becomes

P∝T

It means pressure of the gas is directly proportional to its absolute temperature at constant volume which is law of increase of pressure.

Perfect Gas Equation from Kinetic theory of gases

From kinetic theory of gases, we have

P=13MVˉc2

or,V=13MPˉc2…(i)

Since ˉC2∝T so, equation (i) becomes,

PV∝T

or,PV=RT

Where R is the constant foe one mole of an ideal gas. This equation is the perfect gas equation.

Dalton's Law of Partial Pressure from Kinetic theory of gases

Consider a mixture of gases in a vessel of volume V in which no chemical interaction takes place. . If p1, p2, . . .. .. . are the partial pressure of the respective gases, then from the kinetic theory of gases. we have,

P1=13n1m1Vˉc21,P2=13n2m2Vˉc22,…andsoon

and so on

where, n1,n2, n3, . . . . . are the number of molecules of the respective gases,

m1, m2, m3 . . . . . are the mass of molecules of the respective gases,

and c1, c2, c3. . . . . are the root mean square velocity of the respective gases,

According to Dalton's law, the total pressure exerted by a mixture of pressure is equal to this sum of the pressure exerted by each gas. So total pressure

p=p1+p2+p3+…

=13n1m1Vˉc21+13n2m2Vˉc22+⋯+13nnmnVˉc2n

=13V[n1m1c21+n2m2c22+…nnmnc2n]

If the temperature of the mixture of gases is same, their kinetic energy per molecule must be equal.

That is

12m1c21=12m2c22=⋯=12m3c23=12mnc2n

So, we have

p=13Vmˉc2[n1+n2+⋯+nn]

or,p=mˉc2Vn

where, n1,n2, n3, . . . . . is the total numbers of molecules of the mixture of gases.

∴P=13MVˉc2