It is not a closed system, but the surrounding air is more or less at atmospheric pressure, and the exhaust, once out of the pipes, too. Both and their mixture will remain at atmospheric pressure, more or less, at all times. So they do occupy definite volumes.
Let's looks at a very simplified case, where two volumes of gas (exhaust gas and surrounding air) do not mix but just exchange energy.
In that case, in PV=nRT, n is constant and we can write, for each mass of gas, V/T=k with k being constant.
So V1 is the gas out of the exhaust, with temperature T1.
V2 is the surrounding air, with T2.
V3 is the final volume of mass V1, after cooling to T3.
V4 is the final volume of mass V2, after heating to T4.
I'll further assume the masses of each volume of gas are the same (n1=n2) and that then the final temperatures, T3=T4 are simply (T1+T2)/2 (I am not 100% sure that this holds water, but won't be that far from reality). In this case, also, k1=n1RP is equal to k2=n2RP. I'll call that just k.
Then, V1=kT1 and V2=kT2.
V1 V3 V3 2V3 V2 2V4
-- = k = -- = --------- = -----. Similarly, -- = -----
T1 T3 (T1+T2)/2 T1+T2 T2 T1+T2
.....V1(T1+T2) kT1(T1+T2) k(T1+T2) k(T1+T2)
V3 = --------- = ---------- = --------. Similarly, V4 = --------
2T1 2T1 2 2
Total initial volumes = V1+V2 = kT1+kT2
..............................2k(T1+T2)
Total final volumes = V3+V4 = --------- = kT1+kT2 = V1+V2
2
The volume of a mixture of gases at the same temperature (T3=T4) is the same as the sum of their volumes if they had stayed separated.
I suspect that the case can be generalized, but I don't want to spend the time on it.
So there is no such thing as a free lunch, to heat and expand some gas, you have to cool other gas, which contracts. Overall, you are injecting a lot of high speed gas into free air, but after that, there is no net expansion.
Rivals, not enemies. (Now paraphrased from A. Newey).