Question

The U-shaped tube in FIGURE P18.57 has a total length of $1.0m$. It is open at one end, closed at the other, and is initially filled with air at $20°C\mathrm{and}1.0atm$pressure.

Mercury is poured slowly into the open end without letting any air escape, thus compressing the air. This is continued until the open side of the tube is completely filled with mercury. What is the length $L$of the column of mercury?

Short answer

The length of the column of mercury is$\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{}\mathit{m}$

Step-by-step solution

Given values and used formulas 

Given Information;

Total length of U-shaped tube :$1.0m$.

Initially filled with air at localid="1648649527169" $20°C$

Pressure $=1.0atm$

Theory used ;

In thermodynamics, an isothermal process is a type of thermodynamic process in which the temperature $T$of a system remains constant: $\Delta T=0$.

We would be correct in believing that the air trapped inside the tube is undergoing an isothermal process. Isothermal processes, as we all know, are defined by

$p_1{V}_1=p_2{V}_2$

Finding the length of the column of mercury 

The beginning pressure is equal to atmospheric pressure, whereas the initial volume is equal to the volume of the entire tube, $V_1=A·1$, because the tube length is $1$.

The pressure after compression will be equal to the sum of atmospheric pressure and mercury column pressure; that is, $p_2=p_a+\rho gh$, whilst the volume will plainly be $V_2=A(1-L)$.

As a result, we may use the formula for isothermal reactions to calculate $p_{a}A=(p_a+\rho gL)A(1-L)$

It is obvious that this is a second-order equation and that the cross-section areas of the tube can be cancelled.

$p_a=p_a-p_{a}L+\rho gL-\rho g{L}^2$, on the other hand, is what we're left with.

Now one can cancel the two atmospheres because each term contains $L$. Therefore, a much simpler equation is left:

$$\begin{gathered} -p_a+\rho g+\rho gL=0 \\ \Rightarrow \rho gL=\rho g-p_a \\ \Rightarrow L=\frac{\rho g-p_a}{\rho g}L \\ =1-\frac{p_a}{\rho g} \end{gathered}$$

Substituting the numbers,

$$\begin{gathered} L=1-\frac{101300}{13600\times 9.8} \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{}\mathit{m} \end{gathered}$$