By Sunil Bhardwaj

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From the maxwell distribution law, $$\frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -M{ c }^{ 2 } }{ 2RT } \right) }{ c }^{ 2 }dc$$ $$\therefore \frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -E }{ RT } \right) }{ c }^{ 2 }dc\qquad ......(1)$$ where $$\frac { 1 }{ 2 } M{ c }^{ 2 }=E$$ or Kinetic Energy of gas molecules. $$\therefore M{ c }^{ 2 }=2E\qquad ......(2)$$ Takein sq root of eq 2 $${ M }^{ \frac { 1 }{ 2 } }c={ \left( 2E \right) }^{ \frac { 1 }{ 2 } }\qquad ......(3)$$ Also taking derivative of eq 2 $$2Mc dc = 2 dE$$ $$Mc dc = dE\qquad ......(4)$$ Now multiplying eq 3 with 4 $${ M }^{ \frac { 1 }{ 2 } }c \times Mc dc = { \left( 2E \right) }^{ \frac { 1 }{ 2 } } \times dE$$ $${ M }^{ \frac { 3 }{ 2 } }{ c }^{ 2 } dc = { \left( 2E \right) }^{ \frac { 1 }{ 2 } } dE$$ $$\therefore { c }^{ 2 } dc = \frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } } dE \qquad ......(5)$$ Substituting ths value in eq 1, we get $$\frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -E }{ RT } \right) }\frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } } dE$$ $$\therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =4\pi \frac { { \left( M \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 2\pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }\frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } }$$ $$\therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =\frac { { 4\pi } }{ 2\sqrt { 2 } { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \sqrt { 2 } \left( E \right) }^{ \frac { 1 }{ 2 } }$$ $$\therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =\frac { { 2\pi } }{ { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \left( E \right) }^{ \frac { 1 }{ 2 } }$$ $$\therefore p =\frac { { 2\pi } }{ { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \left( E \right) }^{ \frac { 1 }{ 2 } }\qquad ......(6)$$ Where $$p =\frac { 1 }{ n } \frac { { dn }_{ c } }{ dE }$$ represents probability of molecules having energy E

MCQ on Gaseous State from Physical Chemistry
###### Prof. Gianfranco Coletti

Shared publicly - 2019-08-23 00:00:00

Don’t want your columns to simply stack in some grid tiers? Use a combination of different classes for each tier as needed. See the example below for a better idea of how it all works.

###### Prof. Maheshwar Sharon

Shared publicly - 2019-08-24 00:00:00

For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to

###### sunil

Shared publicly - 2023-02-28 11:09:52

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###### ss

Shared publicly - 2023-02-28 10:48:10

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