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I am currently trying to learn some basics of thermodynamics. I encountered the following equation $$ \frac{p_2}{p_1}=\left(\frac{V_1}{V_2}\right)^\gamma $$ that allows computing the effect of an isentropic compression or expansion of an ideal gas from volume $V_1$ to volume $V_2$ on the pressure. $\gamma$ is the heat capacity ratio of the gas.

I tried to apply the formula to the following hypothetic situation: suppose we have two moles of an ideal gas with heat capacity at constant volume $C_v=30~\mathrm{J\cdot K^{-1}\cdot mol^{-1}}$ and heat capacity at constant pressure $C_p=C_v+R$, R being the universal gas constant. The two moles are stored separately at a pressure $p_0 = 1.0\cdot 10^5~\mathrm{Pa}$ and temperature $T_0 = 300~\mathrm{K}$. The volume of each mole can be computed using the ideal gas relation: $V_0=\frac{n\cdot R\cdot T_0}{p_0} \approx 2.49 \cdot 10^{-2}~\mathrm{m^3}$.

We heat the first mole by an energy E = 600 J at constant volume. The first mole now has:

  • A volume $V_1=V_0$
  • A temperature $T_1$ computed using the gas heat capacity at constant volume: $T_1 = T_0 + \frac{E}{C_v \cdot n} = 320~\mathrm{K}$
  • A pressure $p_1$ computed using the ideal gas law: $p_1 = \frac{n \cdot R \cdot T_1}{V_1} \approx 1.07 \cdot 10^{5}~\mathrm{Pa}$.

We heat the second mole by the same amount of energy at constant pressure. The second mole now has:

  • A pressure $p_2 = p_0$
  • A temperature $T_2$ computed using the gas heat capacity at constant pressure: $T_2 = T_0 + \frac{E}{C_p \cdot n} \approx 316~\mathrm{K}$.
  • A volume $V_2$ computed using the ideal gas law: $V_2 = \frac{n \cdot R \cdot T_2}{p_2} \approx 2.62 \cdot 10^{-2}~\mathrm{m^3}$.

We now expand (isentropically) the first mole so it has the same volume as the second mole: $V_1' = V_2$. The new pressure of the first mole $p_1'$ can be computed using the isentropic process equation:

$$ p_1' = p_1 \cdot \left(\frac{V_1}{V_1'}\right)^{\frac{C_p}{C_v}} \approx 99955~\mathrm{Pa} $$

Why isn't $p_1'$ exactly $p_2$? Sorry if I made a silly mistake in my reasoning.

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closed as off-topic by user191954, stafusa, Jon Custer, AccidentalFourierTransform, sammy gerbil Sep 26 '18 at 19:13

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    $\begingroup$ Why would you expect it to be exactly the same? $\endgroup$ – Chet Miller Sep 23 '18 at 13:22
  • $\begingroup$ Because the entropy of the gas heated at constant pressure and the gas heated as constant volume should be the same, and thus an isentropic process modifying the volume of one so it equals the volume of the other should also lead to identical pressure and temperature $\endgroup$ – paly2 Sep 23 '18 at 13:31
  • $\begingroup$ Why do you feel that the entropy of the gas heated at constant pressure should be the same as the gas heated at constant volume? The final temperatures, pressures, and volumes are different. $\endgroup$ – Chet Miller Sep 23 '18 at 13:36
  • $\begingroup$ Because the amount of energy that has been added to the gas by heating it is the same, and the amount of work performed by the gas during expansion in either case --- during or after heating --- is the same $\endgroup$ – paly2 Sep 23 '18 at 13:45
  • $\begingroup$ Or is the amount of work produced by the gas expansion after heating greater than the amount of work produced by the gas expansion during heating? $\endgroup$ – paly2 Sep 23 '18 at 13:53
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The entropy change for the constant volume case is $$nC_v\ln{(T_f/T_i)}=(1)(30)\ln{(320/300)}=1.936\ J/K$$

The entropy change for the constant pressure case is $$nC_p\ln{(T_f/T_i)}=(1)(38.314)\ln{(315.66/300)}=1.950\ J/K$$

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Entropy change doesnot only depend on the heat added.It also depends on the temperature of the system at which heat is added(Temperature might be continuously changing as heat is being added).The work done during adiabatic expansion is not equal to the work done during isobaric expansion(even if the change in volume during expansion is the same).

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