Question

A liquid occupies half of a vessel at a particular temperature. The volume of the unoccupied part remains constant at all temperatures. If \[\alpha \] and \[\gamma \] are the coefficients of linear and real expansions of a vessel and liquid, then \[\gamma = \]
A. \[3\alpha \]
B. \[3\alpha /2\]
C. \[6\alpha \]
D. \[9\alpha \]

Answer 1

Hint: We have given that the volume of the unoccupied part of the vessel remains constant. This implies that the difference in the volume of the vessel and volume of liquid remains constant at any temperatures. Use the expression for volume expansion with change in temperature and express the volume of the unoccupied part of the vessel.

Formula used:
Volume expansion is given as,
\[{V_f} = {V_i}\left( {1 + 3\alpha \Delta T} \right)\]
Here, \[{V_f}\] is the final volume, \[{V_i}\] is the initial volume, \[\alpha \] is the coefficient of linear expansion and \[\Delta T\] is the temperature change.

Complete step by step solution:
We assume at temperature \[{T_1}\], the volume of the vessel is \[{V_V}\] and volume of liquid is \[{V_L}\].
We have given that, \[{V_L} = \dfrac{{{V_V}}}{2}\].
Since the volume of unoccupied part remains constant at any temperatures, we can write,
\[{V_V} - {V_L} = {\text{constant}}\] …… (1)
We know that the coefficient of volume expansion is three times the coefficient of linear expansion. Therefore, we have, \[{\alpha _V} = 3{\alpha _L}\], where, \[{\alpha _L}\] is the coefficient of linear expansion.
Assuming the volume of the vessel at temperature \[{T_2}\]is \[{V'_V}\] and volume of liquid is \[{V'_L}\].
 We can express the volume expansion of vessel with temperature as follows,
\[{V'_V} = {V_V}\left( {1 + 3\alpha \Delta T} \right)\] …… (2)
Also, the volume expansion of liquid is,
\[{V'_L} = {V_L}\left( {1 + \gamma \Delta T} \right)\] …… (3)
The volume of the unoccupied part of the vessel at temperature \[{T_2}\] is \[{V'_V} - {V'_L}\] . We have given that the volume of the unoccupied part remains constant at any temperature. Therefore, we can write,
\[{V'_V} - {V'_L} = {V_V} - {V_L}\]
Using equation (2) and (3), we have,
\[{V_V}\left( {1 + 3\alpha \Delta T} \right) - {V_L}\left( {1 + \gamma \Delta T} \right) = {V_V} - {V_L}\]
\[ \Rightarrow {V_V} + 3\alpha \Delta T{V_V} - {V_L} + \gamma \Delta T{V_L} = {V_V} - {V_L}\]
\[ \Rightarrow 3\alpha {V_V} = \gamma {V_L}\]
Since \[{V_L} = \dfrac{{{V_V}}}{2}\], we can rewrite the above equation as,
\[3\alpha {V_V} = \gamma \dfrac{{{V_V}}}{2}\]
\[ \Rightarrow \gamma = 6\alpha \]

So, the correct answer is “Option C”.

Note:
If the volume coefficient of expansion is not given and you know the linear coefficient of expansion, then you can use \[{\alpha _V} = 3{\alpha _L}\], where, \[{\alpha _L}\] is the coefficient of linear expansion. Since the volume of the unoccupied part of the vessel remains constant, there is simultaneous increase in the volume of the vessel and decrease in the volume of liquid.