critical volume of a gas is three times the value of the vandaer Waals constant.true or false
Answers
Answered by
0
Explanation:
A Vander Waals equation is given by:
(P+
V
2
a
)(V−b)=RT
Where, a and b are constant
Solving above equation, P=
V−b
RT
−
V
2
a
Taking derivative of P w.r.t volume
∂V
∂P
=0
∂V
2
∂
2
P
=0
So, P becomes:
∂V
∂P
=−
(V−b)
2
RT
+
V
3
2a
=0
V
3
2a
=
(V−b)
2
RT
...........(1)
V
4
a
=
2V(V−b)
2
RT
..............(2)
Taking double derivative again,
∂V
2
∂
2
P
=
(V−b)
3
2RT
−
V
4
6a
=0
or
(V−b)
3
RT
=
V
4
3a
Put equation (2) in above equation
(V−b)
3
RT
=
2V(V−b)
2
3RT
On rearranging,
3V−3b=2V
V
c
=3b
Vc is critical volume
Use this value in equation (1)
4b
2
RT
=
27b
3
2a
T
c
=
27Rb
8a
Tc
Answered by
0
Answer:
It is false
hope it helps u
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