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In cylinder with movable piston a gas is closed. It has been noticed that if cylinder was thermally isolated, quasi-statistical growth of volume resulted in reduction of pressure according to equation: $$p^{3} V^{5} = const$$ Find work done over this system and heat transfered to it for three different processes presented on the graph(ADB, ACB and linear AB). Graph 1

My attempts to solve this exercise:

I found a first law of thermodynamics equation: $$dU = dQ - dW,$$ where $dW = - pdV$ and $dQ = TdS$ (I guess). I am also aware of existing an ideal gas law: $$pV =nRT.$$ I think I could calculate $dW$ easily (with information given on the graph), then use the depency of pressure and volume (given by first equation) and with the help of ideal gas law - find a temperature, but what with $dS$ then? It's necessary to find the heat transfer $dQ$. I will be glad for every help.

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To get the change in temperature in each case, you use $$\Delta T=\frac{\Delta (PV)}{nR}$$ To get the change in internal energy in each case, you use $$\Delta U=nC_v\Delta T$$ To get $C_v$, you use $$C_v=\frac{3}{2}R$$ To get the work, you use $$W=\int{PdV}$$To get the heat, you use $$Q=\Delta U+W$$

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  • $\begingroup$ But is it possible to find n in this equations? There aren’t number of moles and type of gas given exercise $\endgroup$ – Sebastian Kupisiewicz Jan 3 '18 at 9:06
  • $\begingroup$ If you do it algebraically, the n cancels out. $\endgroup$ – Chet Miller Jan 3 '18 at 11:57
  • $\begingroup$ To the person who gave this a down vote: Which part of the analysis were you not able to follow? $\endgroup$ – Chet Miller Jan 3 '18 at 12:39
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Work is the area under the graph P-V. So you just need to integrate (there're lots of triangles, use them).

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  • $\begingroup$ So for example the work done would be: $$W = \fraq{1}{2} \cdot ( * \cdot 10^{-3} - 10^{-3} ) \cdot (10^{5} - \fraq{1}{32} \cdot 10^{5} ) \approx 339 J,$$ $$ W_{ACB} = - W_{ADB} \approx -339J,$$ and what about AB? $\endgroup$ – Sebastian Kupisiewicz Jan 2 '18 at 19:05
  • $\begingroup$ Edit your comment, in order to get it in latex, change "fraq" for "frac" (Otherwise, it's realliy diffult to read). $\endgroup$ – MatMorPau22 Jan 2 '18 at 19:10
  • $\begingroup$ I did it one time( also incorrect) and I can't do it again(newbie's things) so I will just try to rewrite my comment: So for example the work done would be: $$W_{ADB} = \frac{1}{2} \cdot ( 8 \cdot 10^{-3} - 10^{-3} ) \cdot (10^{5} - \frac{1}{32} \cdot 10^{5} ) \approx 339 J,$$ $$ W_{ACB} = - W_{ADB} \approx -339J,$$ and what about AB? Does $W_{AB}$ equal $W_{ACB}$ (same area)? Thanks for the hint! $\endgroup$ – Sebastian Kupisiewicz Jan 2 '18 at 19:18
  • $\begingroup$ Sorry for the hint, isn't a hundred per hundred correct. From $A$ to $B$ is suposed to evole according to $P^3V^5=K_1$. However, you can determine this constant, and then define the curve $P(V)=\frac{K_2}{V^{\frac{5}{3}}}$ and integrate (it isn't a triangle now). The work done from $A$ to $B$ deppends on the path, however you can find integrating (and plotting triangles), assuming that your base level is $P=0$ $\endgroup$ – MatMorPau22 Jan 2 '18 at 19:25
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    $\begingroup$ Ok, so the work done in AB process would be $$W_{AB} = \int_{10^{3}}^{8 \cdot 10^{-3}} \frac{K_{2}}{v^{\frac{5}{3}}}dv ?$$ $\endgroup$ – Sebastian Kupisiewicz Jan 2 '18 at 19:36

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