91影视

Concrete table weight $=560kg$

Diameter of the cylinder $=25cm$

Initial lengthrole="math" localid="1648018532150" $=1.20m$

Pressure$=1.0atm$

It is clear from the text that this is an isothermal process.

The work in an isothermal process is given by,

$W=-nRT\ln \left(\frac{V_f}{V_i}\right)$

What we need to find is the volume ratio, the number of moles, and the temperature Since this is an isothermal process, we can write that$p_1{V}_1=p_2{V}_2鈬�\frac{V_2}{V_1}=\frac{p_1}{p_2}$

As for the pressures, on the other side, we know that $p_1=p_a=1.01鈰�{10}^6\mathrm{Pa}$while the second pressure will be,

$p_2=p_1+\frac{F}{A}$

The added term is the pressure coming from the weight distributed on the four cylinders with total cross-section area $A$. Knowing the diameter $D$of all these cylinders and the fact that there are 4 such cylinders, we can write that

$p_2=p_1+\frac{mg}{4\frac{蟺D^2}{4}}鈬�p_2=p_1+\frac{mg}{蟺D^2}$

This means that we now know the volume ratio to be

$\frac{V_2}{V_1}=\frac{p_a}{p_a+\frac{mg}{蟺D^2}}$

As for $n$, let's express this by the Ideal Gas Law: we'll see that this will solve us our problem.

$pV=nRT鈬�n=\frac{pV}{RT}$

Substituting in the expression for the work, we have

$W=-\frac{pV}{RT}RT\ln \left(\frac{p_a}{p_a+\frac{mg}{蟺D^2}}\right)$

Substituting, and taking note that we just need to put values of the pressure and volume at the beginning from the same instant -- let this be the initial values, we have

$W=-p_a蟺D^2{h}_1\ln \left(\frac{p_a}{p_a+\frac{mg}{蟺D^2}}\right)$

Substituting our numerical values, we have

$W=(L鈭�1.01鈰�{10}^{5}Pa鈰�蟺鈰�{0.25}^{2}m鈰�1.20m\ln )\left(\frac{1.01鈰�{10}^{5}Pa}{1.01鈰�{10}^{5}Pa+\frac{560kg鈰�9.8g}{蟺鈰�{0.25}^{2}m}}\right)$

Multiply the expression,

$W=5800\mathrm{J}$