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?

  • $\begingroup$ What is your solution? What do you think you might be doing wrong? $\endgroup$ Commented Mar 21, 2020 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
    Commented Mar 22, 2020 at 11:30
  • $\begingroup$ Please show how you got the 5/6. I get $\frac{4}{3}T_1$ as the first term. $\endgroup$ Commented Mar 22, 2020 at 12:24
  • $\begingroup$ Well, I meant the solution of my textbook, not mine. $\endgroup$
    – Quaerendo
    Commented Mar 22, 2020 at 13:29
  • $\begingroup$ Well, then, I respectfully disagree with your book. However, I concur with the Q term in their solution. $\endgroup$ Commented Mar 22, 2020 at 14:19

1 Answer 1


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.

  • $\begingroup$ 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
    Commented Mar 23, 2020 at 9:35
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 23, 2020 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.