Consider difference of pressures as weight of piston [closed]








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Let us say there is a container with two ideal gases in it divided by a piston.Volume of ideal below the piston is 2L and above is 8L.Total volume of container is 10L.
n of both is 1 mole and T is 300 for both.
Find the Temp when the ratio between the volume of gases is 4L and 6L.


How I solved it :


$\frac{R\*1\*300}{2L}$ - $\frac{R\*1\*300}{8L}$ since the gas below has more pressure.


Now my sir told me that difference in pressures is equal to weight of piston.The equation that I wrote above.My sir told me this because He said it is important into know that difference in pressures is equal to weight of piston .How is that possible?
Please explain


Please do share if there is any other way to solve this if possible.




We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure:


$$\vec F\_g + \vec F\_\mathrm{p,down} + \vec F\_\mathrm{p,up}=\vec 0 \tag{1}$$


If $V$ is the given bottom gas volume, $V\_0$ is the total gas volume, $n$ is the molar amount of each of gases, the gas constant $R \approx \pu{8.314 JK^-1mol^-1}$, $T$ is absolute temperature. $m, g, A$ are piston mass, gravitational acceleration and piston area, then:


$$ m \cdot g + p\_\mathrm{top} \cdot A =p\_\mathrm{p,bottom}\cdot A\tag{2}$$


$$ p\_\mathrm{bottom} - p\_\mathrm{p,top} = \frac{m \cdot g}{A}\tag{3}$$


As we know from the ideal gas state equation that


$$p=\frac{nRT}{V}\tag{4}$$


then the pressure difference between the bottom and upper gases is


$$\Delta p = nRT ( \frac 1{V} - \frac 1{V\_0-V})=\frac {mg}{A}\tag{5}$$


As the right side is constant, we can eliminate it, writing


$$nRT\_1 ( \frac 1{V\_1} - \frac 1{V\_0-V\_1}) = nRT\_2 ( \frac 1{V\_2} - \frac 1{V\_0-V\_2})\tag{6}$$


$$T\_2 = T\_1 \cdot \frac {\frac 1{V\_1} - \frac 1{V\_0-V\_1}} {\frac 1{V\_2} - \frac 1{V\_0-V\_2}}\tag{7}$$


where $T\_1, V\_1$ are the initial temperature and the volume of the bottom gas,$T\_2, V\_2$ are the final temperature and the volume of the bottom gas.


Now we finally plug in the numbers. Doing it before would be just distraction.


$$T\_2 = \ce{300 K} \cdot \frac {\frac 1{\pu{2 L}} - \frac 1{\pu{8 L}}} {\frac 1{\pu{4 L}} - \frac 1{\pu{6 L}}} = \ce{300 K} \cdot \frac {3/8\ \pu{L^-1}} {1/12\ \pu{L^-1}}= \ce{300 K} \cdot \frac 92 = 1350 K\tag{8}$$


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