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Let's suppose we have $N$ molecules of a diatomic ideal gas at temperature $T_1$ and pressure $P_1$ inside a constant volume $V$. The system is isolated from the surroundings so that the internal energy remains constant. The diatomic gas reacts decomposing into a monoatomic one, according to this reaction: $A_2\rightarrow2A+Q$, where $Q$ is the heat released in the reaction (which is exothermic). I need to find out the expression of the final temperature of the system, $T_2$, once all the molecules of the diatomic gas have reacted.

I know that for ideal gases $U=nc_VT$, with $U$ the internal energy of the gas and $c_V$ its molar heat capacity, and that $c_V$ is different for diatomic and monoatomic gases. However, by combining these formulas I don't get the right solution.

How could this be solved?

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  • $\begingroup$ What is your solution? What do you think you might be doing wrong? $\endgroup$ – sammy gerbil Mar 21 at 22:45
  • $\begingroup$ My solution is $T_2=\frac{5}{6}T_1+\frac{Q}{3R}$, where $Q$ is the heat released in each reaction of a mol of $A_2$ into 2 mols of $A$. $\endgroup$ – Quaerendo Mar 22 at 11:30
  • $\begingroup$ Please show how you got the 5/6. I get $\frac{4}{3}T_1$ as the first term. $\endgroup$ – Chet Miller Mar 22 at 12:24
  • $\begingroup$ Well, I meant the solution of my textbook, not mine. $\endgroup$ – Quaerendo Mar 22 at 13:29
  • $\begingroup$ Well, then, I respectfully disagree with your book. However, I concur with the Q term in their solution. $\endgroup$ – Chet Miller Mar 22 at 14:19
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The heat of reaction of an ideal gas, $\Delta H_R(T_0)$, is defined as the change in enthalpy in going from pure reactants to pure products, holding the temperature $T_0$ constant by adding an amount of heat Q equal to $Q=\Delta H_R(T_0)$. So, for an exothermic reaction, $\Delta H_R(T_0)$ is negative. For the same change from reactants to products at constant $T_0$, the change in internal energy is $$\Delta U_R(T_0)=\Delta H_R(T_0)-\Delta (PV)=\Delta H_R(T_0)-(\Delta n)RT_0$$In the present case, $\Delta n=1\ mole$.

For the case of a reaction carried out adiabatically at constant volume, we need to apply Hess' law to get the change in temperature. This leads to $$\Delta U=\Delta U_R(T_0)+n_pC_{vp}(T-T_0)=0$$ where $n_P$ is the number of moles of product in the reaction formula and $C_{vp}$ is the weighted average molar heat at capacity at constant volume of the product mixture. So, in the present case, $n_p=2$ and $C_{vp}$ is the molar heat capacity at constant volume of A. So, combining the above two equations, we would have: $$2C_{vA}(T-T_0)=-\Delta H_R(T_0)+RT_0$$ This development depends on the fact that, for an ideal gas, both U and H are independent of pressure and specific volume.

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  • $\begingroup$ In this case, the volume remains constant, but not the pressure. I understand that enthalpy is defined for processes at constant pressure. Is it still okay to use enthalpy in this case? $\endgroup$ – Quaerendo Mar 23 at 9:35
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    $\begingroup$ Unfortunately, you understand incorrectly. Enthalpy is simply defined as H = U + PV, independent of any process. And since, for an ideal gas, both U and PV are functions only of temperature, for and ideal gas, H is a function only of temperature. $\endgroup$ – Chet Miller Mar 23 at 11:52

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