Question

When a liquid, filled in two vessels $A$ and $B$ of equal volumes, is heated, the coefficient of apparent expansions of the liquids are formed to be ${\gamma _1}$ and ${\gamma _2}$ respectively. If ${\alpha _1}$ be the coefficient of the linear expansion of $A$, then the coefficient of linear expansion of $B$ will be:
A) $\dfrac{{{\gamma _2} - {\gamma _1}}}{3} + {\alpha _1}$
B) $\dfrac{{{\gamma _2} - {\gamma _1}}}{3} - {\alpha _1}$
C) $\dfrac{{{\gamma _1} - {\gamma _2}}}{3} + {\alpha _1}$
D) $\dfrac{{{\gamma _1} - {\gamma _2}}}{3} - {\alpha _1}$

Answer 1

The coefficient of the linear expansion of $B$ can be determined by the using the actual coefficient of the expansion of the liquid for two liquids, then by using this two coefficients equation in the linear expansion equation, the coefficient of linear expansion of $B$ can be determined.

Complete step by step solution:
Given that,
The coefficient of the apparent expansion of the two liquids are, ${\gamma _1}$ and ${\gamma _2}$.
The coefficient of the linear expansion of $A$ is, ${\alpha _1}$.
Assume that the actual coefficients of the expansion of the two liquids are ${\gamma _1}'$ and ${\gamma _2}'$.
Now the actual coefficients of the expansion of the liquid is given by,
${\gamma _1}' = {\gamma _1} - {\gamma _{vessel}}\,..................\left( 1 \right)$
Similarly,
${\gamma _2}' = {\gamma _2} - {\gamma _{vessel}}\,.................\left( 2 \right)$
Now subtracting the equation (2) by equation (1), then
${\gamma _2}' - {\gamma _1}' = {\gamma _2} - {\gamma _{vessel}} - {\gamma _1} + {\gamma _{vessel}}$
By cancelling the same terms with different sign in the above equation, then the above equation is written as,
${\gamma _2}' - {\gamma _1}' = {\gamma _2} - {\gamma _1}\,.....................\left( 3 \right)$
Now, the relation between the coefficient of the linear expansion and the coefficient of the expansion of the liquid is given by,
${\alpha _1}' = \dfrac{{{\gamma _1}'}}{3}$
Where, ${\alpha _1}'$ is the actual linear coefficient.
Now, the linear coefficient
${\alpha _2} = \dfrac{{{\gamma _2}'}}{3} - {\alpha _1}' + {\alpha _1}$
By substituting the ${\alpha _1}'$ value in the above equation, then the above equation is written as,
${\alpha _2} = \dfrac{{{\gamma _2}'}}{3} - \dfrac{{{\gamma _1}'}}{3} + {\alpha _1}$
By rearranging the terms in the above equation, then the above equation is written as,
${\alpha _2} = \dfrac{{{\gamma _2}' - {\gamma _1}'}}{3} + {\alpha _1}$
By substituting the equation (3) in the above equation, then the above equation is written as,
${\alpha _2} = \dfrac{{{\gamma _2} - {\gamma _1}}}{3} + {\alpha _1}$

Hence, the option (A) is the correct answer.

Note: The coefficient of the linear expansion of the $B$ is equal to the sum of the difference of the coefficient of the expansion of the liquids which is divided by the three and the coefficient of the linear expansion of the $A$.