Thermodynamics Problems! [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 4 years ago.








[Improve this question](/posts/7840/edit) 




> 
> A cylinder, 0.5 m long and of cross section $S$, closed at each end is fitted with a smooth thermally insulating piston, so as to contain gas in each end A and B. Initially, the piston is in equilibrium at a distance of 0.2 m from one end of the cylinder, and the gas in each compartment is at temperature 27 °C. The temperature of the smaller volume of gas A is now increased to 177 °C, while that of gas in the larger volume B is maintained at 27 °C. The piston moves through a distance of $X$ until a position is obtained.
> 
> 
> 1) Write an equation relating the initial and final pressure, volume and temperature of the gas in A.
> 
> 
> 2) Write an equation relating the initial and final pressure, volume and temperature of the gas in B.
> 
> 
> 3) Hence find the new position of equilibrium.
> 
> 
> 


P.S. This is not my assignment or homework so you could feel free to explain.




The volume $V$ of a cylinder is given as $V=\pi r^2 h$ where $h$ is the height and $r$ the radius. Since the cross-section of the cylinder is $S$, the radius is given as $r=S/2$.


Now, we know that the piston is at equilibrium in the initial position, and the equilibrium in this case is $p\_\text{A} = p\_\text{B}$ for the pressures of the corresponding gases.


We also know the ideal gas law, which, when adding the information above (and assuming the gases A and B behave ideally) is: $$pV = nRT \Longleftrightarrow p = \frac{nRT}{V} = \frac{nRT}{\pi (S/2)^2 x} = \frac{4nRT}{\pi S^2 x}$$
For the distance $x$ of the piston to one wall.


For gas A we have the initial and final conditions:
$$ p\_\text{i}^{(\text{A})} = \frac{4n^{(\text{A})} RT\_\text{i}}{\pi S^2 x\_\text{i}} \; \text{and} \;\; p\_\text{f}^{(\text{A})} = \frac{4n^{(\text{A})} RT\_\text{f}}{\pi S^2 x\_\text{f}} $$


And for B (with a total height of the cylinder of $h$):
$$ p\_\text{i}^{(\text{B})} = \frac{4n^{(\text{B})} RT\_\text{i}}{\pi S^2 (h-x\_\text{i})} \; \text{and} \;\; p\_\text{f}^{(\text{B})} = \frac{4n^{(\text{B})} RT\_\text{i}}{\pi S^2 (h-x\_\text{f})} $$


We know that the initial and final pressures were the same (equilibrium condition):
$$ \frac{4n^{(\text{A})} RT\_\text{i}}{\pi S^2 x\_\text{i}} = \frac{4n^{(\text{B})} RT\_\text{i}}{\pi S^2 (h-x\_\text{i})} \Longleftrightarrow \frac{n^{\text{(A)}}}{x\_\text{i}} = \frac{n^{\text{(B)}}}{h - x\_\text{i}}$$
$$ \frac{4n^{(\text{A})} RT\_\text{f}}{\pi S^2 x\_\text{f}} = \frac{4n^{(\text{B})} RT\_\text{i}}{\pi S^2 (h-x\_\text{f})} \Longleftrightarrow \frac{n^{\text{(A)}}T\_\text{f}}{x\_\text{f}} = \frac{n^{\text{(B)}}T\_\text{i}}{h-x\_\text{f}}$$


We can now solve both equations for $\frac{n^\text{(A)}}{n^\text{(B)}}$ and equate them:
$$ \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x\_\text{i}}{h-x\_\text{i}} \; \text{and} \;\; \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x\_\text{f}T\_\text{i}}{T\_\text{f}(h-x\_\text{f})} $$
$$ \frac{x\_\text{i}}{h-x\_\text{i}} = \frac{x\_\text{f}T\_\text{i}}{T\_\text{f}(h-x\_\text{f})}$$


Solving for $x\_\text{f}$ gives us the final equation:
$$ x\_\text{f} = \frac{\frac{hx\_\text{i}T\_\text{f}}{T\_\text{i}(h-x\_\text{i})}}{1 + \frac{x\_\text{i}T\_\text{f}}{T\_\text{i}(h-x\_\text{i})}}$$


For the values provided by you I get $x\_\text{f} = 0.25~\mathrm{m}$, which is exactly in the middle.


 6