Answer
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Hint: As we know that surface tension acts upon the topmost layer of any liquid and it is temperature dependent. Therefore, it will be zero when the surface of separation between two phases (here, the two phases are liquid and vapour) disappear.
Complete step by step answer:
We know that the molecules present at the surface experience a net inward pull due to the presence of a larger number of molecules in the liquid side than that on the vapour side. Hence, there is a tendency for the surface molecules to go into the liquid. This tension was later termed as surface tension.
Mathematically, we can say that- “Surface tension is the force acting along the surface of a liquid at right angle to any line of unit length.”
As we know, in a liquid the forces of attraction between the molecules decrease with the increasing temperature, therefore it follows that surface temperature decreases with temperature too. We can explain it in the terms of the following equation:
\[\gamma {{({{M}_{V}})}^{\dfrac{2}{3}}}=a-kt\]
Where, $\gamma $ = surface tension
${{M}_{v}}$ = Molar volume of the liquid
‘a’ and ‘k’ are constants, t= temperature , ${{t}_{c}}$= critical temperature
Critical temperature is the temperature at which there is no separation of surfaces between the vapour phase and the liquid phase. Therefore, at critical temperature the surface tension will become zero. Therefore,
$\gamma {{({{M}_{v}})}^{\dfrac{2}{3}}}=0$
So we can write that, $a-k{{t}_{c}}=0$
Or, $a=k{{t}_{c}}$
And according to Ramsay and Shields Modification, the above equation was modified to get better results. The modification made to the surface tension equation was-
$\gamma {{({{M}_{v}})}^{\dfrac{2}{3}}}=k({{t}_{c}}-t-6)$
Where, the temperature was written in terms of the critical temperature and according to this equation, surface energy becomes zero when,
$\begin{align}
& if,\gamma =0 \\
& then,k({{t}_{c}}-t-6)=0 \\
& or,t={{t}_{c}}-6 \\
\end{align}$
Therefore, as we can see the above derivation states that the surface tension is zero at a temperature ${{6}^{\circ }}C$ below the critical temperature for any liquid. However, it was experimentally seen that for many liquids surface tension was zero at critical temperature or 1 or 2 degrees below the critical temperature.
Therefore, it can be said that the correct option is [C] critical point.
Note: It is important to remember here that surface tension is the force that separates the liquid and the gas phase. Therefore, when the phases do not exist, the force also ceases to exist. Therefore, at critical temperature, it becomes zero.
Complete step by step answer:
We know that the molecules present at the surface experience a net inward pull due to the presence of a larger number of molecules in the liquid side than that on the vapour side. Hence, there is a tendency for the surface molecules to go into the liquid. This tension was later termed as surface tension.
Mathematically, we can say that- “Surface tension is the force acting along the surface of a liquid at right angle to any line of unit length.”
As we know, in a liquid the forces of attraction between the molecules decrease with the increasing temperature, therefore it follows that surface temperature decreases with temperature too. We can explain it in the terms of the following equation:
\[\gamma {{({{M}_{V}})}^{\dfrac{2}{3}}}=a-kt\]
Where, $\gamma $ = surface tension
${{M}_{v}}$ = Molar volume of the liquid
‘a’ and ‘k’ are constants, t= temperature , ${{t}_{c}}$= critical temperature
Critical temperature is the temperature at which there is no separation of surfaces between the vapour phase and the liquid phase. Therefore, at critical temperature the surface tension will become zero. Therefore,
$\gamma {{({{M}_{v}})}^{\dfrac{2}{3}}}=0$
So we can write that, $a-k{{t}_{c}}=0$
Or, $a=k{{t}_{c}}$
And according to Ramsay and Shields Modification, the above equation was modified to get better results. The modification made to the surface tension equation was-
$\gamma {{({{M}_{v}})}^{\dfrac{2}{3}}}=k({{t}_{c}}-t-6)$
Where, the temperature was written in terms of the critical temperature and according to this equation, surface energy becomes zero when,
$\begin{align}
& if,\gamma =0 \\
& then,k({{t}_{c}}-t-6)=0 \\
& or,t={{t}_{c}}-6 \\
\end{align}$
Therefore, as we can see the above derivation states that the surface tension is zero at a temperature ${{6}^{\circ }}C$ below the critical temperature for any liquid. However, it was experimentally seen that for many liquids surface tension was zero at critical temperature or 1 or 2 degrees below the critical temperature.
Therefore, it can be said that the correct option is [C] critical point.
Note: It is important to remember here that surface tension is the force that separates the liquid and the gas phase. Therefore, when the phases do not exist, the force also ceases to exist. Therefore, at critical temperature, it becomes zero.
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