Thermodynamics: Can a reaction be adiabatic and isothermal at the same time?

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MCAT Tachycardia

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So I was going through a online review about adiabatic processes and isothermal processes and was taught that when you have an isothermal reaction and volume goes up then temperature goes up and when volume goes down, temperature goes down. Now with adiabatic, its inverted. If volume goes up, temp goes down. If volume goes down, temp goes up.
So from this I inferred that you can't have both an adiabatic and isothermal reaction at the same time. However, I'm not sure that I got my facts straight, or if I can make this inference. So if anyone knows, please shed some light and let me know. Thanks!

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I'm a little confused by the first part of your statement: "that when you have an isothermal reaction and volume goes up then temperature goes up". Isothermal means that temperature does not change, so by definition, if you have an isothermal reaction, then temperature cannot go up.
But anyway, you cannot have a reaction that does PV work that is also simultaneously adiabatic and isothermal. ΔE = q + w. Adiabatic means no heat flow, so q = 0. ΔE = (0) + w = w. Isothermal means no change in temperature, so ΔT = 0. (Internal) energy is basically a measure of temperature and E∝ T (TPR pg. 215), so if ΔΤ = 0, then ΔE = 0. We already have ΔE = w, and so now we can substitute for ΔΕ, 0 = w.
So the only way you can have a reaction be adiabatic (q = 0) and isothermal (ΔT = 0 and so ΔE = 0) is if there is no work, ie, no expansion or compression of the gas.
Thinking about the whole thing in words might make more sense. For a gas in an insulated container (q = 0) to expand, the kinetic energy of the molecules is used to do the work of raising the piston; the gas molecules are doing work and their own energy is falling. As their energy falls, so also does their temperature fall. When work is being done on the gas (ie, the piston is pushed down), energy is being transferred into the system and the energy & temperature of the molecules increases. On the other hand, for a gas in a container that allows heat flow so that the gas does not change temperature (ΔT = 0 and ΔE = 0), as the gas molecules do work on the piston to expand their volume, they need more heat to come in to replace the energy they are losing by doing work (q is +). As work is being done on the gas molecules to compress the volume, the energy and temperature of the gas molecules would be rising, but heat flows out of the gas (q is -). So you can have no heat flow and a change in temperature during expansion & compression, or you can have heat flow and no change in temperature during expansion & compression. The only way to have no heat flow and no change in temperature is if the gas is not expanding or compressing.
I hope that's all correct! Thermodynamics is pretty confusing.
 
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