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

Questions Related to van der-waal equation: equation of state for real gas

The size of container B is double that of A and gas in B is at double the temperature and pressure than that in A. The ratio of molecules in the two containers will then be -

real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

  1. $\frac{N _B}{N _A} = \frac{1}{1}$

  2. $\frac{N _B}{N _A} = \frac{2}{1}$

  3. $\frac{N _B}{N _A} = \frac{4}{1}$

  4. $\frac{N _B}{N _A} = \frac{1}{2}$

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Correct Option: A

Two vertical parallel glass plates are partially submerged in water. The distance between the plates is $d = 0.10 mm$, and their width is $l  = 12 cm$. Assuming that the water between the plates does not reach the upper edges of the plates and that the wetting is complete, find the force of their mutual attraction.

real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

  1. $17N$

  2. $13N$

  3. $10$

  4. $19N$

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Correct Option: B

For gaseous decomposition of ${PCI} _{5}$ in a closed vessel the degree of dissociation '$\alpha $', equilibrium pressure 'P' & ${'K} _{p}'$ are related as

real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

  1. $\ \alpha =\sqrt { \frac { { K } _{ p } }{ P } } $

  2. $\ \alpha =\frac { 1 }{ \sqrt { { K } _{ p }+P } } $

  3. $\ \alpha =\sqrt { \frac { { K } _{ p }+P }{ { K } _{ p } } } $

  4. $\alpha =\sqrt { { K } _{ p }+P } $

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Correct Option: A
Explanation:

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

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

 

If pressure of ${CO} _{2}$ (real gas) in a container is given by $P=\cfrac { RT }{ 2V-b } -\cfrac { a }{ 4{ b }^{ 2 } } $, then mass of the gas in container is:

real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

  1. $11g$

  2. $22g$

  3. $33g$

  4. $44g$

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Correct Option: B
Explanation:

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


$\bigg( P +\dfrac{n^2 a}{V^2}\bigg)(V- nb)=nRT\implies P=\dfrac{nRT}{V-nb}-\dfrac{n^2a}{V^2}$

Given that Pressure of $CO _2$ gas in a contaner is given by:
$P= \dfrac{RT}{2V-b}-\dfrac{a}{4b^2}$

Compairing it with the standard Van der waal's equation we get :
$n=\dfrac12$

Therefore, Number of moles in a container , $n=\dfrac12$
Molar mass of $CO _2= 44\ gm$
Mass of gas in the container, $m= \dfrac12\times 44 =22 gm$