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Question: A glass flask contains some mercury at room temperature. It is found that at different temperatures ...

A glass flask contains some mercury at room temperature. It is found that at different temperatures the volume of air inside the flask remains the same. If the volume of mercury in flask is 300cm3300c{m^3}, then the volume of flask is (given that the coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8×10−4(0C)−11.8 \times {10^{ - 4}}{{(^0}C)^{ - 1}} and 9×10−6(0C)−19 \times {10^{ - 6}}{{(^0}C)^{ - 1}} respectively):
A) 4500cm34500c{m^3}
B) 450cm3450c{m^3}
C) 2000cm32000c{m^3}
D) 6000cm36000c{m^3}

Explanation

Solution

Hint:- We know that volume of a matter changes with change in its temperature. In given examples the volume of mercury and flask change with change in their temperature. Here the volume of the flask is the sum of the volume of mercury and air inside the flask. Volume of air remains same then change in volume of glass is equal to change in volume of flask.

Complete step by step solution:
Given, the volume of mercury is VHg=300cm3{V_{Hg}} = 300c{m^3}.
Volume expansion of mercury is γHg=  1.8×10−4(0C)−1{\gamma _{Hg}} = \;1.8 \times {10^{ - 4}}{{(^0}C)^{ - 1}}.
Let the volume of the flask is Vflask{V_{flask}}.
Volume expansion of mercury is γflask=  3×9×10−6(0C)−1{\gamma _{flask}} = \;3 \times 9 \times {10^{ - 6}}{{(^0}C)^{ - 1}} (as we know volume expansion coefficient is three time of linear expansion coefficient).
Change in volume is given by ΔV=V×γ×ΔT\Delta V = V \times \gamma \times \Delta T.
Change in volume of mercury is
ΔVHg=300×1.8×10−4×ΔT\Delta {V_{Hg}} = 300 \times 1.8 \times {10^{ - 4}} \times \Delta T -(1)
Change in volume of flask is
ΔVflask=Vflask×3×9×10−6×ΔT\Delta {V_{flask}} = {V_{flask}} \times 3 \times 9 \times {10^{ - 6}} \times \Delta T -(2)
Here the volume of the flask is the sum of the volume of mercury and air inside the flask. Volume of air remains same then change in volume of glass is equal to change in volume of flask.
Hence, ΔVflask=ΔVHg\Delta {V_{flask}} = \Delta {V_{Hg}} -(3)
From equation (1), (2) and (3), we have
Vflask×3×10−6=300×1.8×10−4{V_{flask}} \times 3 \times {10^{ - 6}} = 300 \times 1.8 \times {10^{ - 4}}
Vflask=300×1.8×10−43×9×10−6=2000cm3{V_{flask}} = \dfrac{{300 \times 1.8 \times {{10}^{ - 4}}}}{{3 \times 9 \times {{10}^{ - 6}}}} = 2000c{m^3}

Hence the correct answer is option C.

Note: For a matter coefficient of volume expansion is three times of linear expansion of that and surface coefficient is two times of linear coefficient. Sometimes temperature coefficient also changes with change in temperature. Mercury is used in thermometers due to its constant temperature coefficient.