In Vander Waal's equation the critical $P_{c}$ is given by

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a}{27 b^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{27 a}{b^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{b^{2}}{a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{a}{27 b^{2}}$
The Vander Wall's equation of state is
$(P + \frac{a}{V^{2}})$ $(V - b)$ = $R T$
$P = \frac{R T}{V - b} - \frac{a}{V^{2}}$
At the critical point,
$P = P_{C}, V = V_{C}$ and $T = T_{C}$
$∴ P_{C} = \frac{R T_{C}}{V_{C} - b} - \frac{a}{V_{C}^{2}}$ .......(i)
At the critical point on the isothermal,
$\frac{d P_{C}}{d V_{C}} = 0$
$∴ 0 = \frac{- R T_{C}}{(V_{C} - b)^{2}} + \frac{1 a}{V_{C}^{3}}$
$\frac{R T_{C}}{(V_{C} - b)^{2}} = \frac{2 a}{V_{C}^{3}}$ .... (ii)
Also at critical point,
$\frac{d^{2} P_{C}}{d V_{C}^{2}} = 0$
$0 = \frac{2 R T}{(V_{C} - b)^{3}} - \frac{6 a}{V_{C}^{4}}$
$\frac{2 R T_{C}}{(V_{C - b})^{3}}$ = $\frac{5 a}{V_{C}^{4}}$ ....(iii)
Dividing (ii) by (iii) we get
$\frac{1}{2} (V_{C} - b)$ = $\frac{1}{3} V_{C}$
$V_{C} = 3 b$ ..... (iv)
Putting this value in (ii), we get
$\frac{R T_{C}}{4 b^{2}} = \frac{2 a}{27 b^{3}}$
$T_{C} = \frac{8 a}{27 b R}$ .......(v)
Putting the value of $V_{C}$ and $T_{C}$ in (i), we get
$P_{C}$ = $\frac{R}{2 b} (\frac{8 a}{27 b R})$ = $\frac{a}{9 b^{2}}$
$= \frac{a}{27 b^{2}}$

Suggest Corrections

Similar questions

V_{c} = 3 b, P_{c} = a / 27 b^{2}, T_{c} = 8 a / 27 b R

Z

P_{c} =

T_{c} =

V_{c} =

[π = \frac{P}{P_{c}}, ϕ = \frac{V}{V_{c}} a n d θ = \frac{T}{T_{c}}]

a

a = 2.51 L^{2} a t m {m o l e}^{- 2}

b = 0.032 L {m o l e}^{- 1}

= 30

0.40 g . c m^{3}

T_{c} = \frac{2 \times 10^{5}}{821} K

a

L^{2} m o l^{- 2})

Join BYJU'S Learning Program