
During the evaporation of cleaned sugarcane juice, pressure surrounding it is reduced to:
A. Increase the boiling point
B. Maintain the boiling point at 373K
C. Decrease the boiling point
D. Increase the size of the sugar crystals
Answer
446.7k+ views
Hint:The variation of vapour pressure of a liquid with temperature is given by the Clausius Clapeyron's equation, pressure is indirectly proportional to boiling point.
Complete step-by-step answer:In order to answer our question, we need to learn about vapour pressure of liquid solutions. When a liquid is taken in a beaker covered from above at certain temperature, a part of the liquid evaporates and its vapours fill the space available to them. As the evaporation proceeds, the number of molecules in vapour phase gradually increases and they move about randomly in a space above the liquid some of these molecules strike the liquid surface and get condensed. Very soon a state of an equilibrium will be established between liquid and vapour phases. The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid. The variation of vapour pressure of a liquid with temperature is given by the Clausius Clapeyron's equation:
\[\log \dfrac{{{p}_{2}}}{{{p}_{1}}}=\dfrac{{{\Delta }_{vap}}H}{2.303R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]\]
Here, ${{p}_{1}},{{p}_{2}}$ are pressures at temperature ${{T}_{1}},{{T}_{2}}$ respectively.${{\Delta }_{vap}}H$ is the evaporation enthalpy.
So, as pressure is decreased, the boiling point of the liquid also decreases.
So, we get the final answer for this question as option C.
Note: The vapour pressure of a liquid depends on the following factors:
A. Nature of the liquid: Liquids, which have weak intermolecular forces, are more volatile because their molecules have greater tendency to escape into vapour phase. Consequently, they have larger vapour pressure.
B. Temperature: This is due to the reason that with increase in temperature the kinetic energy of molecules increases which further increases their tendency to vaporise.
Complete step-by-step answer:In order to answer our question, we need to learn about vapour pressure of liquid solutions. When a liquid is taken in a beaker covered from above at certain temperature, a part of the liquid evaporates and its vapours fill the space available to them. As the evaporation proceeds, the number of molecules in vapour phase gradually increases and they move about randomly in a space above the liquid some of these molecules strike the liquid surface and get condensed. Very soon a state of an equilibrium will be established between liquid and vapour phases. The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid. The variation of vapour pressure of a liquid with temperature is given by the Clausius Clapeyron's equation:
\[\log \dfrac{{{p}_{2}}}{{{p}_{1}}}=\dfrac{{{\Delta }_{vap}}H}{2.303R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]\]
Here, ${{p}_{1}},{{p}_{2}}$ are pressures at temperature ${{T}_{1}},{{T}_{2}}$ respectively.${{\Delta }_{vap}}H$ is the evaporation enthalpy.
So, as pressure is decreased, the boiling point of the liquid also decreases.
So, we get the final answer for this question as option C.
Note: The vapour pressure of a liquid depends on the following factors:
A. Nature of the liquid: Liquids, which have weak intermolecular forces, are more volatile because their molecules have greater tendency to escape into vapour phase. Consequently, they have larger vapour pressure.
B. Temperature: This is due to the reason that with increase in temperature the kinetic energy of molecules increases which further increases their tendency to vaporise.
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