8. The unit of vander waal's constants ‘a' and 'b'are respectively
1) L atm? mol-1 and mol L-1
2) L atm mol2 and mol L
3) L2 atm mol-2 and mol-IL
4) L-2 atm-' mol-1 and Lit mol-2
Answers
Answer:
Unit of a is atm.dm^{ 6 }.mo{ l }^{ -2 }atm.dm
6
.mol
−2
and unit of 'b' is { dm }^{ 3 }.dm
3
. {mol }^{ -1 }mol
−1
The pressure produced by the attractive force is proportional to the square of the number density.
The Vander Waal's equation is given below,
\left( P\quad +\quad \frac { a }{ { V }^{ 2 } } \right) (V\quad -\quad b)\quad =\quad RT(P+
V
2
a
)(V−b)=RT
P = Pressure
V = Volume
T = Temperature
'a' and 'b' are constants.
The\quad unit\quad of\quad 'a':Theunitof
′
a
′
:
P\quad +\quad a\frac { { n }^{ 2 } }{ { V }^{ 2 } }P+a
V
2
n
2
a\quad =\quad \frac { P{ V }^{ 2 } }{ { n }^{ 2 } }a=
n
2
PV
2
a\quad =\quad \frac { atm.(d{ m }^{ 3 }{ ) }^{ 2 } }{ (mol{ ) }^{ 2 } }a=
(mol)
2
atm.(dm
3
)
2
a\quad =\quad atm.\quad dm^{ 6 }.\quad mo{ l }^{ -2 }a=atm.dm
6
.mol
−2
The\quad unit\quad of\quad 'b':Theunitof
′
b
′
:
\frac { V }{ n } \quad -\quad b
n
V
−b
b\quad =\quad \frac { V }{ n }b=
n
V
b\quad =\quad { dm }^{ 3 }{ .mol }^{ -1 }b=dm
3
.mol
−1
Answer:
3)L2 atm mol-2 and mol-IL
Explanation: