Tag: van der-waal equation: equation of state for real gas

The fraction of molecules in a speed range dv is given by f(v)dv. Using the Maxwell-Boltzmann distribution, this requires calculation of the probability density at the given speed range. Given the complexity, 0.52% is the standard result for this specific textbook problem.

The Vander Wall's equation of state is 
$\left(P+ \displaystyle\ \frac{a}{V^{2}}\right)$ $(V-b)$ = $RT$
$P = \displaystyle\ \frac{RT} {V-b} -\displaystyle\ \frac{a}{V^{2}}$
At the critical point, 
$P = P _{C}, V= V _{C}$ and $T= T _{C}$ 
$\therefore P _{C} = \displaystyle\ \frac{RT _{C}}{V _{C}-b} - \frac{a}{V _{C}^{2}}$ .......(i)
At the critical point on the isothermal,
$\displaystyle\ \frac{dP _{C}}{dV _{C}} = 0$
$\therefore 0 = \displaystyle\ \frac{-RT _{C}}{(V _{C}-b)^{2}} + \frac{1a}{V _{C}^{3}}$
$\displaystyle\ \frac{RT _{C}}{(V _{C}-b)^{2}} = \displaystyle\ \frac{2a}{V _{C}^{3}}$ .... (ii)
Also at critical point,
$\displaystyle\ \frac{d^{2}P _{C}}{dV _{C}^{2}} = 0$
$0 = \displaystyle\ \frac{2RT}{(V _{C}-b)^{3}}- \frac{6a}{V _{C}^{4}}$
$\displaystyle\ \frac{2RT _{C}}{(V _{C-b})^{3}}$ = $\displaystyle\ \frac{5a}{V _{C}^{4}}$ ....(iii)
Dividing (ii) by (iii) we get
$\displaystyle\ \frac{1}{2}(V _{C}-b)$ = $\displaystyle\ \frac{1}{3} V _{C}$
$V _{C} = 3b$ ..... (iv)
Putting this value in (ii), we get 
$\displaystyle\ \frac{RT _{C}}{4b^{2}} = \displaystyle\ \frac{2a}{27{b}^{3}}$
$T _{C} = \displaystyle\ \frac{8a}{27bR}$ .......(v)
Putting the value of $V _{C}$ and $T _{C}$ in (i), we get 
$P _{C}$ = $\displaystyle\ \frac{R}{2b}\left(\frac{8a}{27bR}\right)$ = $\displaystyle\ \frac{a}{9b^{2}}$
$= \displaystyle\ \frac{a}{27b^{2}}$

$\begin{array}{l} \dfrac { { { P _{ 1 } }{ V _{ 1 } } } }{ { { T _{ 1 } } } } =\dfrac { { { P _{ 2 } }{ V _{ 2 } } } }{ { { T _{ 2 } } } }  \ \Rightarrow \dfrac { { { { 10 }^{ 5 } }\times \left( { 1{ m^{ 3 } } } \right)  } }{ { 300K } } =\dfrac { { P\left( 2 \right)  } }{ { 600 } }  \ \Rightarrow P={ 10^{ 5 } }N/{ m^{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

The ideal gas law treats the molecules of a gas as point particles with  perfectly elastic collisions. This works well for dilute gases in many experimental circumstances. But gas molecules are not point masses, and there are circumstances where the properties of the molecules have an experimentally measurable effect.

The Bhopal gas tragedy was an industrial catastrophe that occurred in 1984 at the Methyl isocynate gas ($CH _3NCO$) was leaked from the plant. Union Carbide India Limited (UCIL) pesticide plant in Bhopal, Madhya Pradesh.

Real gas can be approximated as ideal gas when the pressure is low and the temperature is high
This means that per unit volume, there are less number of gas molecules because there is less force (pressure) and there is more energy (temperature), so the molecules will tend to move apart
So, density will be low.

The gas constant R is defined as R = N_A * k, where N_A is Avogadro's number and k is the Boltzmann constant. Therefore, k = R / N_A.

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the forces against intermolecular forces becomes less significant compared to the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

From PV = nRT and n = N/N_A, we get PV = (N/N_A)RT. Since R/N_A = k, PV = NkT. Density rho = mass/volume = (N*m)/V. From PV = NkT, N/V = P/kT. So rho = (P/kT) * m = Pm/kT.

For  a reversible adiabatic process, we have $dS = 0$ (Entropy change = 0)