Real Gases and Kinetic Theory

Using Charles's Law (V1/T1 = V2/T2) at constant pressure, if V2 = 3V1, then T2 = 3T1. Given T1 = 0 C = 273 K, T2 = 3 * 273 = 819 K. Converting back to Celsius, 819 - 273 = 546 C.

This is a free expansion (Joule expansion) into a vacuum. Since the walls are insulating (adiabatic) and no work is done (expansion against vacuum), the internal energy remains constant, meaning temperature T remains constant. The volume doubles, so by PV = nRT, the pressure must halve to P/2.

Volume V = 554 = 100 m^3. At STP, 1 mole occupies 22.4 liters (0.0224 m^3). Number of moles n = 100 / 0.0224 approx 4464 moles. Number of molecules = n * Avogadro's number = 4464 * 6e23 approx 2.68e27, which is of the order of 10^27.

PV=RT is ideal gas equation and gases behave ideally only at high temperature and low pressure.
Therefore option(B).

$\left( {p + \frac{a}{{{V _2}}}} \right)\left( {v - b} \right) = cT$

For adiabatic processes, P * V^gamma = constant. For a diatomic gas, gamma = 1.4 or 7/5. P2 = P1 * (V1/V2)^gamma. Here V1/V2 = 32. P2 = P * (32)^(7/5) = P * (2^5)^(7/5) = P * 2^7 = 128P.

The rate of collisions per unit area is proportional to (n * v_avg), where n is number density and v_avg is average speed. Since n = N/V and V, T are same, n is same. v_avg is proportional to 1/sqrt(M). Ratio = v_He / v_O2 = sqrt(M_O2 / M_He) = sqrt(32 / 4) = sqrt(8) = 2*sqrt(2).

Real gases deviate most from ideal behavior at high pressures and low temperatures, where intermolecular forces and molecular volume become significant.

In Langmuir's model of adsorption of a gas on solids surfaces. The adsorption at a single site on the surface may invoice multiple molecules at the ame time. The mass of gas striking at a given area of surface is independent of the pressure of the gas.

At low pressure and high temperature real gas obey PV=RT  i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas

Given $P=50 \ atm , \ \ V=350 ml =0.350 \ \ \ litre, \ \ n=1 \ \ mole $ and $ T=300 K$

At high temperature and low pressure the real gas behaves as an ideal gas.

Van der Walls equation takes into account inter atomic forces between gas particles which is not considered in the ideal gas model. Since simplicity of  liquification of a gas depends upon forces between its particles, Van der Walls equation is followed by easily liquifieable gases. 

No. of moles, $n=\dfrac{m}{M}=\dfrac{2}{4}=0.5 mol$

$\displaystyle b = 4N \times \frac{4}{3} \pi \frac{d^3}{8}$ (standard definition of boyles constant)

$\displaystyle = \frac{4 \times 6.02 \times 10^{23} \times 3.14 \times 2.94^3 \times 10^{-30}}{3 \times 8}$

$\displaystyle = 32 \times 10^{-6}$

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.

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)

${  \quad \quad \quad \quad \quad \quad \quad PCl } _{ 5 }\rightleftharpoons { PCl } _{ 3(9) }+{ Cl } _{ 2(9) }\\ Initial\quad mole\quad \quad \quad 1\quad  \quad 0\quad\quad\quad 0\\ After\quad mole\quad \quad 1-\alpha \quad \quad  \alpha \quad\quad  \alpha \\ decomposition$

Total mole$=1-\alpha+\alpha+\alpha\\=1+\alpha$

Total pressure$=P$

Partial pressure of $PCl _5=P(\cfrac{1-\alpha}{1+\alpha})$

PArtial pressure of $PCl _3=P(\cfrac{\alpha}{1+\alpha})$

Partial pressure of $PCl _2=P(\cfrac{\alpha}{1+\alpha})$

Then $K _p=\cfrac{(PCl _3)(Cl _2)}{(PCl _5)}\\ \quad=\cfrac{P(\cfrac{\alpha}{1+\alpha})P(\cfrac{\alpha}{1+\alpha})}{P(\cfrac{1-\alpha}{1+\alpha})}\\ \quad=\cfrac{P^2\alpha^2}{(1+\alpha)^2}\times\cfrac{(1+\alpha)}{P(1-\alpha)}\\K _p=\cfrac{P\alpha^2}{1-\alpha^2}$

now, $1-\alpha^2<<1$

so that $K _p=P\alpha^2\\ \alpha^2=\cfrac{K _p}{P}\\ \alpha=\sqrt{\cfrac{K _p}{P}}$

According to Van Der waal's equation for $n$ mole of real gas