Question

A one litre flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. The volume of mercury taken in the flask is : (Coefficient of linear expansion of glass is $ 9 \times {10^{ - 6}}{}^o{C^{ - 1}} $ and coefficient of volume expansion of $ Hg $ is $ 1.80 \times {10^4}{}^o{C^{ - 1}} $ )
(A) $ 150ml $
(B) $ 750ml $
(C) $ 1000ml $
(D) $ 700ml $

Answer 1

Hint :Use the formula for volume expansion of solid to find the volume of mercury taken in the flask. THE formula for volume expansion is given by, $ \Delta V = V\gamma \Delta \theta $ , where $ \Delta V $ is the change in volume $ V $ is the initial volume $ \gamma $ is the volume expansion coefficient and $ \Delta \theta $ is the temperature change of the material in which the volume expansion of the material happens.

Complete Step By Step Answer:
We know that the volume expansion of a material is given by, $ \Delta V = V\gamma \Delta \theta $ , where $ \Delta V $ is the change in volume $ V $ is the initial volume $ \gamma $ is the volume expansion coefficient and $ \Delta \theta $ is the temperature change of the material in which the volume expansion of the material happens.
So, for glass we can write, $ {\left( {\Delta V} \right)_g} = V{\gamma _g}\Delta \theta $ . Now, we have given the coefficient of linear expansion of glass $ {\alpha _g} $ . We know that the volume expansion coefficient is related to linear expansion coefficient as $ \gamma = 3\alpha $ . Hence, we can write, $ {\left( {\Delta V} \right)_g} = V{\gamma _g}\Delta \theta = V(3{\alpha _g})\Delta \theta $ .
For mercury the change of volume of mercury can also be written similarly. So, $ {\left( {\Delta V} \right)_m} = {V_m}{\gamma _m}\Delta \theta $ .
Now, since volume of the air does not change due to the change of temperature, hence the change of volume of flask must be equal to the change of volume of mercury hence we can write,
 $ {\left( {\Delta V} \right)_g} = {\left( {\Delta V} \right)_m} $
Or, $ 3{\alpha _g}V = {\gamma _m}{V_m} $
Putting the values, coefficient of linear expansion of flask $ {\alpha _g} = 9 \times {10^{ - 6}}{}^o{C^{ - 1}} $ , coefficient of volume expansion of mercury $ {\gamma _m} = 1.80 \times {10^4}{}^o{C^{ - 1}} $ and the volume of flask $ V = 1l $ we get,
 $ 3 \times 9 \times {10^{ - 6}} \times 1 = 1.80 \times {10^4}{V_m} $
On simplifying we get ,
 $ {V_m} = \dfrac{{27 \times {{10}^{ - 6}}}}{{1.80 \times {{10}^4}}} $
Or, $ {V_m} = 0.15 $
Hence, volume of mercury taken in flask is $ 0.15l $ or $ 0.15 \times 1000 = 150ml $
Hence, option (A ) is correct.

Note :
 $ \bullet $ The temperature change for both the flask and mercury is the same.
 $ \bullet $ When the volume of air changes then we can find the change in volume of the air using Charles’s law of ideal gas and then find the volume of the mercury in the flask.
 $ \bullet $ The coefficient of areal expansion is related to linear expansion as $ \beta = 2\alpha $ where $ \beta $ is the coefficient of areal expansion.