Solveeit Logo
Login

Question

Question: A glass flask of volume one liter at 0°C is filled level full of mercury at this temperature. The fl...

A glass flask of volume one liter at 0°C is filled level full of mercury at this temperature. The flask and mercury are now heated to 100°C. How much mercury will spill out, if the coefficient of volume expansion of mercury is 1.82×10−41.82\times {{10}^{-4}}/°C and linear expansion of glass is 0.1×10−40.1\times {{10}^{-4}}/°C, respectively?

A. 2.48×10−2L2.48\times {{10}^{-2}}L
B. 1.52×10−2L1.52\times {{10}^{-2}}L
C. 1.53×104L1.53\times {{10}^{4}}L
D. 1.52×10−4L1.52\times {{10}^{-4}}L

Explanation

Solution

Hint: Consider the volume expansion of mercury and glass flask separately. Also take the increment of radius into consideration. Volume varies with the third power of length. As a result, the volume expansion coefficient is thrice the value of length expansion coefficient.

Formula used:
Volume of the flask is given by,
V=πr2lV=\pi {{r}^{2}}l……………………….(1)
Where, r is the radius of the flask
And l is the height of the flask.
Volume expansion due to temperature is,
ΔV=Vγ(Δθ)\Delta V=V\gamma (\Delta \theta )……………………(2)

Where,
VV is the initial volume.
γ\gamma is the volume expansion coefficient
Δθ\Delta \theta is the change in temperature

Complete step by step answer:
We will approach the problem by considering the volume expansion of mercury and glass flask separately.
First, Equation (2) gives us the change in volume of the mercury,
ΔV1=(1L)(1.82×10−4)(100−0)\Delta {{V}_{1}}=(1L)(1.82\times {{10}^{-4}})(100-0)
⇒ΔV1=1.82×10−2\Rightarrow \Delta {{V}_{1}}=1.82\times {{10}^{-2}}
So, the volume of the mercury will increase by,
1.82×10−21.82\times {{10}^{-2}}L
Now, the volume of the flask depends on the radius and the height of the flask. As the linear expansion coefficient is extremely small with respect to the size of the flask, we can make the following approximation,
ΔV2≈V(3α)(Δθ)\Delta {{V}_{2}}\approx V(3\alpha )(\Delta \theta )

Where,
V is the volume of the flask
α\alpha is the linear expansion coefficient
Δθ\Delta \theta is the change in temperature.
The factor of 3 comes because volume varies with the third power of length.
So, the change of volume of the flask is,
ΔV2≈V(3α)(Δθ)\Delta {{V}_{2}}\approx V(3\alpha )(\Delta \theta )
=(1L)(3×0.1×10−4)(100−0)=(1L)(3\times 0.1\times {{10}^{-4}})(100-0)
=0.3×10−2=0.3\times {{10}^{-2}}
Hence, the mercury spilled in the process,
ΔV1−ΔV2\Delta {{V}_{1}}-\Delta {{V}_{2}}
=1.82×10−2−0.3×10−2=1.82\times {{10}^{-2}}-0.3\times {{10}^{-2}}
=1.52×10−2=1.52\times {{10}^{-2}}

So, the correct answer is (B).

Note: You can determine the true dependency by using differentiation. It will give a more accurate answer. However, the values are so small that there won’t be much of a difference. You need to find the volume expansions separately. You should not incorporate both of them at the same time for the sake of simplicity.