Question

The values of the van der Waal’s constants for a gas are a = 4.10 $dm^{6}$ bar $mol^{-2}$ and b = 0.035 $dm^{3}$ $mol^{-1}$. Calculate the values of the critical temperature and critical pressure for the gas.

Answer 1

i) Calculation of critical temperature = $4.16 d{ m }^{ 6 }bar.mo{ l }^{ -2 }$

$b = 0.035d{ m }^{ 3 }bar.mo{ l }^{ -1 }$

$R = 0.0821 d{ m }^{ 3 }bar.mo{ l }^{ -1 }{ K }^{ -1 }$

Now, critical temperature, ${ T }_{ c }\quad =\quad \frac { 8a }{ 27Rb }$

Substituting the values${ T }_{ c }=\frac { 8\times4.10 }{ 27\times0.0821\times 0.035 } =422.76K$

ii) Calculation of critical pressure

${ P }_{ e }=\frac { a }{ 27{ b }^{ 2 } }$

Substitute the values

$= \frac { 4.10 }{ 27 \times 0.035 \times 0.035 } = 123.96 bar$

Therefore, critical temperature is 422.76 K and critical pressure is 123.96 bar.

Answer 2

i) Calculation of critical temperature = 4.16 d{ m }^{ 6 }bar.mo{ l }^{ -2 }4.16dm

6

bar.mol

−2

b = 0.035d{ m }^{ 3 }bar.mo{ l }^{ -1 }b=0.035dm

3

bar.mol

−1

R = 0.0821 d{ m }^{ 3 }bar.mo{ l }^{ -1 }{ K }^{ -1 }R=0.0821dm

3

bar.mol

−1

K

−1

Now, critical temperature, { T }_{ c }\quad =\quad \frac { 8a }{ 27Rb }T

c

=

27Rb

8a

Substituting the values{ T }_{ c }=\frac { 8\times4.10 }{ 27\times0.0821\times 0.035 } =422.76KT

c

=

27×0.0821×0.035

8×4.10

=422.76K

ii) Calculation of critical pressure

{ P }_{ e }=\frac { a }{ 27{ b }^{ 2 } }P

e

=

27b

2

a

Substitute the values

= \frac { 4.10 }{ 27 \times 0.035 \times 0.035 } = 123.96 bar=

27×0.035×0.035

4.10

=123.96bar

Therefore, critical temperature is 422.76 K and critical pressure is 123.96 bar.