
The Clausius-Clapeyron equation may be given as:
${\log _{10}}\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\Delta {H_{vap}}}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} A.\dfrac{1}{{{T_2}}}} \right]$
B.$\dfrac{{dP}}{{dT}} = \dfrac{{\Delta S}}{{\Delta V}}$
C.$\dfrac{{dP}}{{dT}} = \dfrac{q}{{T\Delta V}}$
D.$\dfrac{{dP}}{{dT}} = \dfrac{{\Delta V}}{{\Delta S}}$
Answer
559.2k+ views
Hint: We have studied that Claus-Clapeyron relation is a distinct way of characterizing a discontinuous phase transition between two phases of matter of a single constituent. We must have to remember that the Claus-Clapeyron equation basically gives the relation between the pressure and the latent heat of the constituent.
Complete step by step answer:
The equation has been derived and mathematically represented using a graph of pressure-temperature. On the graph there is a line that separates the two phases of the constituent which is known as the coexistence curve.
From the graph, the equation can be deduced as
$\dfrac{{dP}}{{dT}} = \dfrac{L}{{T\Delta V}} = \dfrac{q}{{T\Delta V}} = \dfrac{{\Delta S}}{{\Delta V}}$
From ideal gas approximation it can be written as,
${\log _{10}}\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\Delta {H_{vap}}}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
Where, $\dfrac{{dP}}{{dT}}$ is the slope of the tangent, L & q is the specific latent heat, delta V is the volume change and delta S is the entropy change of phase transition, T is the temperature of the system.
We must remember that the ideal gas equation of this form is very useful as it gives the direct relation of equilibrium or saturation vapour pressure and temperature to the latent heat of the phase change.
Therefore, the correct answer to the question option A, B and C.
Note:
This equation has many applications especially in chemistry. In chemistry and chemical engineering, it is used for transition between a gas and a condensed phase with specific approximation. It is used in meteorology and climatology as well. The atmospheric water vapour drives many important meteorological phenomena, raising the interest of dynamics. It is also useful for evaluation of heat sublimation of ice.
Complete step by step answer:
The equation has been derived and mathematically represented using a graph of pressure-temperature. On the graph there is a line that separates the two phases of the constituent which is known as the coexistence curve.
From the graph, the equation can be deduced as
$\dfrac{{dP}}{{dT}} = \dfrac{L}{{T\Delta V}} = \dfrac{q}{{T\Delta V}} = \dfrac{{\Delta S}}{{\Delta V}}$
From ideal gas approximation it can be written as,
${\log _{10}}\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\Delta {H_{vap}}}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
Where, $\dfrac{{dP}}{{dT}}$ is the slope of the tangent, L & q is the specific latent heat, delta V is the volume change and delta S is the entropy change of phase transition, T is the temperature of the system.
We must remember that the ideal gas equation of this form is very useful as it gives the direct relation of equilibrium or saturation vapour pressure and temperature to the latent heat of the phase change.
Therefore, the correct answer to the question option A, B and C.
Note:
This equation has many applications especially in chemistry. In chemistry and chemical engineering, it is used for transition between a gas and a condensed phase with specific approximation. It is used in meteorology and climatology as well. The atmospheric water vapour drives many important meteorological phenomena, raising the interest of dynamics. It is also useful for evaluation of heat sublimation of ice.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

