Find $ΔU$ for the combustion of methane in a sealed rigid adiabatic container.

Solution says since $q = 0$ and $w = 0,$ so $ΔU = 0$ by the first law of thermodynamics.

Why isn't the heat released during combustion considered?

Does the temperature of the system changes during the process? I know that

$$ΔU = nC_VΔT$$

is not valid directly as the amount of substance is changing, and so do the degrees of freedom.

  • 1
    $\begingroup$ The answer to you first doubt is that the temperature of the products is higher than the temperature of the reactants in adiabatic combustion. That's where the "heat" goes. Your 2nd doubt answers itself. $\endgroup$ Feb 27, 2021 at 13:05
  • $\begingroup$ Sir if temperature is increasing , the way I am calculating ΔU is by ( Uₚ - Uᵣ ) but after writing the expression with help of U = fnRT/2 (Where f = degree of freedoms) , I am not getting ΔU = 0 , please tell where I am going wrong whether this method of calculating ΔU is wrong or something else ? $\endgroup$ Feb 27, 2021 at 16:44
  • $\begingroup$ Do you know how to determine $\Delta U$ if the final temperature were equal to the initial temperature (taking into account that there have been changes in the amounts of chemical species present)? Would you know $\Delta H$ for the case where the initial and final temperatures are equal? $\endgroup$ Feb 27, 2021 at 17:01
  • $\begingroup$ I guess ΔH could be written with help of bond enthalpy , and ΔU is where I am confused ,according to my previous logics ΔU should come out to be zero. $\endgroup$ Feb 27, 2021 at 17:26
  • $\begingroup$ $\Delta H$ is the heat of combustion which you look up or calculate from tabulated heats of formation. Do you know how to do that? If you don't know how to use this to determine $\Delta U$ at constant temperature for the reaction, you need to research that. This is in every thermo book. Once you have that, you need to determine what the temperature rise of the products would have to be in order for the overall $\Delta U$ to be zero. $\endgroup$ Feb 27, 2021 at 19:38

1 Answer 1


Yes, the temperature changes. The 1st law says that the internal energy of an isolated system is constant, but it doesn't prevent it from changing form. When you combust methane, you change the chemical composition, such that you go from bonds with a higher total potential energy to bonds with a lower total potential energy. The lost potential energy is converted to thermal energy, causing the temperature to increase.

  • $\begingroup$ Can you please also clarify how "1st law says that internal energy of a isolated system is constant " , doesn't the heat of reaction increase the internal energy ? {ΔU = q (from reaction) } $\endgroup$ Feb 27, 2021 at 7:27
  • $\begingroup$ You're thinking q is the thermal energy generated by the reaction. It's not. q is the heat that flows into out of the system, as a result of the thermal energy generated by the rxn. But if you isolate the system so heat can't flow in or out, q must be 0, thus all thermal energy changes caused by the reaction must stay inside the system, which is why the system's temperature changes. If the reaction releases thermal energy, as in your example, T goes up. If it absorbs thermal energy, T goes down. $\endgroup$
    – theorist
    Feb 27, 2021 at 17:03
  • $\begingroup$ Please note the following: meta.stackexchange.com/help/someone-answers $\endgroup$
    – theorist
    Feb 27, 2021 at 18:05
  • $\begingroup$ I got your point that the heat of reaction is ''internal'' and we only include heat in/out from the surroundings in FLOT ; Can you tell will the ΔU will be negative if the walls are made conducting instead of adiabatic as now heat is released out of the system to the surroundings $\endgroup$ Feb 27, 2021 at 18:14
  • $\begingroup$ If there is no non-pV work, then w=0, because the walls are rigid, and hence delta_U= q+w = q. If heat flows out, q<0 and thus delta_U <0. $\endgroup$
    – theorist
    Jul 27, 2021 at 17:14

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