# Change of temperature in the decomposition of an ideal diatomic gas into a monoatomic one at constant volume

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?

• What is your solution? What do you think you might be doing wrong? – sammy gerbil Mar 21 '20 at 22:45
• 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 $A$. – Quaerendo Mar 22 '20 at 11:30
• Please show how you got the 5/6. I get $\frac{4}{3}T_1$ as the first term. – Chet Miller Mar 22 '20 at 12:24
• Well, I meant the solution of my textbook, not mine. – Quaerendo Mar 22 '20 at 13:29
• Well, then, I respectfully disagree with your book. However, I concur with the Q term in their solution. – Chet Miller Mar 22 '20 at 14:19

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.