# Adiabatic compressed air and energy calculations

I was thinking since we can assume that all of the energy of the hot gas is transferred to the steel. Wouldn't what you said before still work?

haruspex
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I was thinking since we can assume that all of the energy of the hot gas is transferred to the steel. Wouldn't what you said before still work?
Will all of the gained internal energy will be transferred? Even when the gas cools back to 18C it will be at a higher pressure than it started. That represents a gain in energy, no?

That's true. but this question allows me to state assumptions

is there anything i can assume in order to make this true?

Could i say that the increase in energy due to an increase in pressure is neglible?

haruspex
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Could i say that the increase in energy due to an increase in pressure is neglible?
I would not have thought so, but I could be quite wrong about this. Maybe the internal energy is just PV, so it all does end up in the steel.
@Chestermiller , you'll be able to settle this immediately, I'm sure.

Chestermiller
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I would not have thought so, but I could be quite wrong about this. Maybe the internal energy is just PV, so it all does end up in the steel.
@Chestermiller , you'll be able to settle this immediately, I'm sure.
In my judgment, the way he did it in the first place is correct. The assumption is that the gas adiabatically heats up first, and then equilibrates thermally with the cylinder. So, the way you originally had it doped out was correct.

haruspex
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the gas adiabatically heats up first, and then equilibrates thermally with the cylinder.
Yes, I understand that, but I had this nagging thought that surely the compressed gas, even after equilibration, is storing energy. But I now think that its potential to do work comes from its low entropy, in the same way that an uneven temperature distribution has the capacity to do work.
Thanks for clearing it up.

Chestermiller
Mentor
Yes, I understand that, but I had this nagging thought that surely the compressed gas, even after equilibration, is storing energy. But I now think that its potential to do work comes from its low entropy, in the same way that an uneven temperature distribution has the capacity to do work.
Thanks for clearing it up.
Rather than assuming the the gas heats up first and then equilibrates with the cylinder, it is also possible to assume that the two heat up simultaneously. In this case $$dU=-PdV$$or $$MCdT=-\frac{nRT}{V}dV$$or equivalently $$MC\frac{dT}{T}=\frac{nR}{V}dV$$where M is the mass of the cylinder and C is the heat capacity of the steel. I think that this is more likely what they had in mind. In this case, the expansion would be nearly isothermal (because of the large thermal inertia of the cylinder).

haruspex