on the basis of Kinetic theory of gases, derive an expression for the pressure exerted by the gas
Answers
Explanation:
Consider a cubical vessel (of side l) with walls perfectly elastic as in the figure. Let the vessel have one gram molecule of a gas with its molecules in random motion.
Consider a molecule (mass m) moving with a velocity c, that can be resolved into three components u,v, and w in the direction of the edges of the cube along x,y,z axes.
∴c
2
=u
2
+v
2
+w
2
........(1)
Let us consider two faces of the cube, say P and Q, normal to x−axis.When the molecule collides with the side Q with a velocity u, it rebounds with −u, while its other components remain unchanged.
Change in momentum of the molecule due to this collision =−2mu, which is imparted to the wall per collision.
The time taken by molecule to cover distance l=
u
l
∴, after every interval of time
u
2l
, the molecule will again collide with the wall Q, and the number of collisions per unit time with the wall Q is equal to
2l
u
.
∴, Momentum imparted to the wall per unit time =2mu×
2l
u
=
l
mu
2
(ignoring the negative sign).
The pressure exerted on the wall Q due to one molecule =
l
mu
2
×
l
2
1
=
l
3
mu
2
If the vessel contains N molecules with velocities u
1
,u
2
,u
3
,....u
n
along x−axis, then the pressure exerted by them on Q, say
P=
l
3
m
(u
1
2
+u
2
2
+u
3
2
+...+u
n
2
)=
v
mN
(
u
ˉ
)
2
where V is the volume of the vessel and (
u
ˉ
)
2
is the average value of u
2
of all N molecules.
(
u
ˉ
)
2
=
N
u
1
2
+u
2
2
+u
3
2
+...+u
n
2
As the molecules move randomly,
(
u
ˉ
)
2
=(
v
ˉ
)
2
=(
w
ˉ
)
2
=
3
(
c
ˉ
)
2
=
where
c
ˉ
=
N
c
1
2
+c
2
2
+c
3
2
+...+c
n
2
.........(2)
c
ˉ
is called the root mean square velocity of the molecules.
∴P=
3
1
V
mN(
c
ˉ
)
2
.......(3)
This equation gives the pressure exerted by the gas on the walls of the vessel.It has the same value in any direction because the molecules have no preference for direction.
We have ρ=
V
mN
∴P=
3
1
ρ(
c
ˉ
)
2
or P=
3
2
E where E=
2
1
ρ(
c
ˉ
)
2
is called the mean kinetic energy per unit volume of the gas.
Answer:
KINETIC THEORY OF GASES.
Explanation:
Consider a cubical vessel (of side l) with walls perfectly elastic as in the figure. Let the vessel have one gram molecule of a gas with its molecules in random motion.
Consider a molecule (mass m) moving with a velocity c, that can be resolved into three components u,v, and w in the direction of the edges of the cube along x,y,z axes.
∴c
2
=u
2
+v
2
+w
2
........(1)
Let us consider two faces of the cube, say P and Q, normal to x−axis.When the molecule collides with the side Q with a velocity u, it rebounds with −u, while its other components remain unchanged.
Change in momentum of the molecule due to this collision =−2mu, which is imparted to the wall per collision.
The time taken by molecule to cover distance l=
u
l
∴, after every interval of time
u
2l
, the molecule will again collide with the wall Q, and the number of collisions per unit time with the wall Q is equal to
2l
u
.
∴, Momentum imparted to the wall per unit time =2mu×
2l
u
=
l
mu
2
(ignoring the negative sign).
The pressure exerted on the wall Q due to one molecule =
l
mu
2
×
l
2
1
=
l
3
mu
2
If the vessel contains N molecules with velocities u
1
,u
2
,u
3
,....u
n
along x−axis, then the pressure exerted by them on Q, say
P=
l
3
m
(u
1
2
+u
2
2
+u
3
2
+...+u
n
2
)=
v
mN
(
u
ˉ
)
2
where V is the volume of the vessel and (
u
ˉ
)
2
is the average value of u
2
of all N molecules.
(
u
ˉ
)
2
=
N
u
1
2
+u
2
2
+u
3
2
+...+u
n
2
As the molecules move randomly,
(
u
ˉ
)
2
=(
v
ˉ
)
2
=(
w
ˉ
)
2
=
3
(
c
ˉ
)
2
=
where
c
ˉ
=
N
c
1
2
+c
2
2
+c
3
2
+...+c
n
2
.........(2)
c
ˉ
is called the root mean square velocity of the molecules.
∴P=
3
1
V
mN(
c
ˉ
)
2
.......(3)
This equation gives the pressure exerted by the gas on the walls of the vessel.It has the same value in any direction because the molecules have no preference for direction.
We have ρ=
V
mN
∴P=
3
1
ρ(
c
ˉ
)
2
or P=
3
2
E where E=
2
1
ρ(
c
ˉ
)
2
is called the mean kinetic energy per unit volume of the gas.