Question

A one litre glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in this flask if coefficient of linear expansion of glass is $9\times {10}^{-6}{/}^{\circ }C$ while of volume expansion of mercury is $1.8\times {10}^{-4}{/}^{\circ }C$

Answer 1

Let initially volume of mercury and ar in the flask be $V_1$ & $V_2$

so, volume of flask,

$V=V_1+V_2\to \left(1\right)$ [ where v = 1L]

after heating by $△T$, the volume of flask and mercury become

$V^{\prime }$ and $V_1^{\prime }$ respectively, while $V_2$ remain same.

$V^{\prime }=V_1^{\prime }+V_2\to \left(2\right)$

From (1) & (2),

$V-V_2=V^{\prime }-V^{\prime }1$

Applying volmetric expansion,

$V-V_1=V(1+r△T)-V_1(1+r_1△T)$

$\Rightarrow Vr△T=V_1{r}_1△T$

Let initially volume of mercury and ar in the flask be $V_1$& $V_2$

so, volume of flask,

$V=V_1+V_2\to \left(1\right)$ [ where v = 1L]

after heating by $△$ T, the volume of flask and mercury become

$V^{\prime }$ and $V^{\prime }$ respectively, while $V_2$ remain same.

$V^{\prime }=V_1^{\prime }+V_2\to \left(2\right)$

From (1) & (2),

$V-V_2=V^{\prime }-V_1^{\prime }$

Applying volmetric expansion,

$V-V_1=V(1+r△T)-V_1(1+r_1△T)$

$\Rightarrow Vr△T=V_1{r}_1△T$

$\Rightarrow V_1=\frac{Vr}{r_1}=\frac{1\times 3\times 9\times {10}^{-6}}{1.8\times {10}^{-4}}$ $[asr=3\alpha ]$

$\Rightarrow V_1=0.15L=150mL.$ (Ans)