Heat released during combustion at constant pressure vs volume

[ChasingZebras]

Greetings! I've been brushing up on some thermodynamics recently and came across a perplexing sentence in my notes and text from undergrad.

It says that for a combustion reaction, such as the combustion of heptane:
C7H16 (l) + 11O2 (g) ---> 7CO2 (g) + 8H2O (l)
That this process carried out at constant pressure would release more energy than the same process carried out at constant volume.

I realize that enthalpy change is: dH = dU + d(PV), and that under constant pressure dH = q, and under constant volume dH = q + V(dP)

dU = q - w
Where q is heat added into the system, and w is work done by the system

Since the combustion is exothermic, q is negative
The moles of gaseous products are less than the moles of gaseous reactants, thus under constant pressure conditions, the volume of the system has decreased, thus work is done ON the system by the surroundings, and w here is negative (making the term positive). Thus,
dU = -q - (-w)
dU = -q + w
And, dH = -q + w + V(dP) + P(dV)
dH = -q + P(dV) + V(0) + P(-dV)
dH = -q

Under constant volume conditions, once products are formed the pressure of the system has decreased (because less moles of gas), and NO work is done since volume is constant. Thus,
dU = -q - (0)
dU = -q
And, dH = -q + V(dP)
If the pressure of the system has decreased, then dP is negative
dH = -q - V(dP)

So we have at constant pressure: dHp = -q
and at constant volume: dHv = -q - V(dP)

Therefore, wouldn't the process carried out at constant volume release more energy than the same process carried out at constant pressure? Since dHv < dHp, thus more exothermic/more energy release?

Did I go wrong somewhere? Or are is there a possible error in my notes/text?

Any clarification is much appreciated. Thank you for your time!

 

[mjc123]

You are right that q is negative, but that doesn't mean dH = -q. dH = q, and dH is also negative.
You seem to be assuming that q is the same for both conditions. Why? What actually is the same?

 

[Chestermiller]

We need to be careful, because we are dealing with two different sets of starting and end states here.

A. Combustion at constant pressure:
Initial state: Reactants at To, Po, Vo
Final State: Products at To, Po, and V1

B. Combustion at constant volume:
Initial state: Reactants at To, Po, Vo
Final State: Products at To, P1, Vo

Note that, in situation B, the final state is different from the final state in situation A. This needs to be taken into account in relating the heats of combustion for the two changes.

Let's first consider situation A. For this situation, we have $$Q_A=\Delta H_A$$ and $$\Delta U_A=\Delta H_A-(P_0V_1-P_0V_0)=\Delta H_A-\Delta n RT_0=Q_A-\Delta n RT_0$$

Next, let's consider situation B. For this situation, we have $$Q_B=\Delta U_B$$

We next need to relate the internal energy changes in situations A and B. Since the initial states are the same in both situations, by Hess' law, we can write: $$\Delta U_B=\Delta U_A+\Delta U_{AB}$$where $\Delta U_{AB}$ is the change in internal energy from the final state of situation A to the final state of situation B. In the final states, the temperatures in the two situations are exactly the same and we are dealing in both cases solely with the products of the reaction. Since the internal energy of an ideal gas products mixture is a function only of temperature, we must have $\Delta U_{AB}=0$. Therefore, it follows from the previous relationships that:
$$\Delta U_B=\Delta U_A$$and $$Q_B=Q_A-\Delta nRT_0$$Therefore, if the change in the number of moles of gas $\Delta n$ is negtive (as in your example), the amount of heat that would have to be removed when carrying out the reaction at constant volume would have to be less than the amount of heat that would have to be removed at constant pressure. This is consistent with what your notes say.

 

[Chris Riccard]

That doesn't make any sense when your talking about energy, higher pressure is just condensing the energy and the more volume the more you have to work with. Why do you figure these two different quanitys bear the same relationship?

 

[Chestermiller]

That doesn't make any sense when your talking about energy, higher pressure is just condensing the energy and the more volume the more you have to work with. Why do you figure these two different quanitys bear the same relationship?

Just because this doesn't make sense to you, doesn't mean that it doesn't make sense. (What you've written doesn't make sense to me. I've never heard of energy being condensed, for example) The analysis I presented is based on a routine straightforward thermodynamic treatment of the problem. If you are having trouble understanding one (or more) of the equations I've presented, maybe I can help. Have you had a course in thermodynamics yet?

 

[Chris Riccard]

No I haven't, my fault. I'm trying to express density of kenetic energy,

 

[ChasingZebras]

We need to be careful, because we are dealing with two different sets of starting and end states here.

A. Combustion at constant pressure:
Initial state: Reactants at To, Po, Vo
Final State: Products at To, Po, and V1

B. Combustion at constant volume:
Initial state: Reactants at To, Po, Vo
Final State: Products at To, P1, Vo

Note that, in situation B, the final state is different from the final state in situation A. This needs to be taken into account in relating the heats of combustion for the two changes.

Let's first consider situation A. For this situation, we have $$Q_A=\Delta H_A$$ and $$\Delta U_A=\Delta H_A-(P_0V_1-P_0V_0)=\Delta H_A-\Delta n RT_0=Q_A-\Delta n RT_0$$

Next, let's consider situation B. For this situation, we have $$Q_B=\Delta U_B$$

We next need to relate the internal energy changes in situations A and B. Since the initial states are the same in both situations, by Hess' law, we can write: $$\Delta U_B=\Delta U_A+\Delta U_{AB}$$where $\Delta U_{AB}$ is the change in internal energy from the final state of situation A to the final state of situation B. In the final states, the temperatures in the two situations are exactly the same and we are dealing in both cases solely with the products of the reaction. Since the internal energy of an ideal gas products mixture is a function only of temperature, we must have $\Delta U_{AB}=0$. Therefore, it follows from the previous relationships that:
$$\Delta U_B=\Delta U_A$$and $$Q_B=Q_A-\Delta nRT_0$$Therefore, if the change in the number of moles of gas $\Delta n$ is negtive (as in your example), the amount of heat that would have to be removed when carrying out the reaction at constant volume would have to be less than the amount of heat that would have to be removed at constant pressure. This is consistent with what your notes say.

Thank you so much for this comprehensive reply - I see where my faulty thinking was. I incorrectly assumed the amount of heat (q) evolved would be the same under both conditions. It all makes sense now ;D