6398 Views
5389 Views
9374 Views
5215 Views
4115 Views
12382 Views
Let us consider a closed system in which a pure liquid water and its vapors are in equilibrium with each other. $$ Water (liquid) \Longleftrightarrow Water (vapor) $$ For this system Clapeyron equation can be written as, $$ \boxed { \frac { dP }{ dT } = \frac { \Delta H }{ T\left( { V }_{ v } - { V }_{ l } \right) } } $$ or $$ \boxed { \frac { dP }{ dT } = \frac { \Delta { L }_{ v } }{ T\left( { V }_{ v } - { V }_{ l } \right) } } $$ Where \({ L }_{ v }\) is latent heat of vaporization. \({ V }_{ v }\) is Volume of one mole of water in vapor phase. \(V_{ l }\) is Volume of one mole of water in liquid phase.
If temerature is not near the critical temerature, \({ V }_{ v }>>{ V }_{ l }\) and hence \({ V }_{ v } - { V }_{ l } \simeq { V }_{ v } \) $$\therefore \frac { dP }{ dT } = \frac { \Delta { L }_{ v } }{ T{ V }_{ v } } \qquad ....(1)$$ Assuming that the vapor obeys ideal gas law, \(P{ V }_{ v } = nRT\) for one mole $$ P{ V }_{ v } = RT i.e. { V }_{ v } = \frac { RT }{ P } \qquad ....(2) $$ Lets substitute it in equation (1) $$ \frac { dP }{ dT } = \frac { \Delta { L }_{ v } }{ T } \frac { P }{ RT } $$ $$ \therefore \boxed { \frac { dP }{ P } = \frac { \Delta { L }_{ v } }{ R } \frac { dT }{ { T }^{ 2 } } } \qquad ....(3) $$ This equation is known as Clapeyron Clausius Equation .
The equation (3) can be integrated between the limits \({ P }_{ 1 }\) to \({ P }_{ 2 }\) and \(T_{ 1 }\) to \({ T }_{ 2 }\) assuming \(\Delta { L }_{ v }\) remains constant for small range of temerature. $$ \int _{ { P }_{ 1 } }^{ { P }_{ 2 } }{ \frac { dP }{ P } } = \frac { \Delta { L }_{ v } }{ R } \int _{ T_{ 1 } }^{ T_{ 2 } }{ \frac { dT }{ { T }^{ 2 } } } $$ i.e. $$ ln\frac { { P }_{ 2 } }{ { P }_{ 1 } } = \frac { -\Delta { L }_{ v } }{ R } \left( \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ T_{ 1 } } \right) $$ i.e. $$ 2.303log\frac { { P }_{ 2 } }{ { P }_{ 1 } } = \frac { \Delta { L }_{ v } }{ R } \left( \frac { 1 }{ T_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right) $$ $$ \boxed { log\frac { { P }_{ 2 } }{ { P }_{ 1 } } = \frac { \Delta { L }_{ v } }{ 2.303R } \left( \frac { { T }_{ 2 } - { T }_{ 1 } }{ T_{ 1 }{ T }_{ 2 } } \right) } $$ This equation is integrated form of Clapeyron Clausius Equation .
5100 Views
4453 Views
5266 Views
4490 Views
5000 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..