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fog37
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- TL;DR Summary
- Internal energy vs Enthalpy
Hello,
In thermodynamics, with systems being represented by gases (can liquids be also included?), the internal energy ##E_{int}## of a system represents the total kinetic energy + the total potential energy of the system: $$E_{int} = KE_{tot}+ PE_{tot}$$ The term ##KE_{tot}=Q## is also called thermal energy (random motion of all internal parts) and is proportional to the system's temperature ##T##. The term ##PE_{tot}## represents the total chemical energy stored in the chemical bonds.
Internal energy ##E_{int}## can be changed via two different mechanisms, which are heat (transfer thermal energy from the system or to the system) and/or mechanical work s done on or by the system: $$\Delta E_{int} = Q - W$$
(I guess absorbing and emitting radiation is a 3rd mechanism that is often not included). Work ##W## is nonzero when the system's shape (volume) is changed. For a gas, work is given by ##W = \int P dV##. The system (gas) has its own internal pressure ##P_{gas}## and there is also an external pressure ##P_{ext}##, generally the constant atmospheric pressure. I guess we can use either pressure in the definition of work ##W##. But the pressure ##P_{gas}## can be variable: I envision a reaction where the internal pressure change with time (ex: explosion).
If the gas is free to expand/contract and change its volume ##V##, work is nonzero and there is a change in energy.
Instead of using internal energy ##E_{int}## and its change ##\Delta E_{int}##, we can clump work with the internal energy into a new state function ##H##, enthalpy, so we don't need to track work and change in system's volume: $$[\Delta E_{int} +W] = Q$$
$$[\Delta E_{int} +W] = \Delta H$$
$$\Delta H = Q$$
which means that enthalpy changes when the system absorbs or emit thermal energy ##Q## via heat (transfer or gain of thermal energy to the environment).
Does the definition of enthalpy ##H## and its changes only apply if the internal and external pressures are both equal and constant? Why? I am missing that important point...
Thanks!
In thermodynamics, with systems being represented by gases (can liquids be also included?), the internal energy ##E_{int}## of a system represents the total kinetic energy + the total potential energy of the system: $$E_{int} = KE_{tot}+ PE_{tot}$$ The term ##KE_{tot}=Q## is also called thermal energy (random motion of all internal parts) and is proportional to the system's temperature ##T##. The term ##PE_{tot}## represents the total chemical energy stored in the chemical bonds.
Internal energy ##E_{int}## can be changed via two different mechanisms, which are heat (transfer thermal energy from the system or to the system) and/or mechanical work s done on or by the system: $$\Delta E_{int} = Q - W$$
(I guess absorbing and emitting radiation is a 3rd mechanism that is often not included). Work ##W## is nonzero when the system's shape (volume) is changed. For a gas, work is given by ##W = \int P dV##. The system (gas) has its own internal pressure ##P_{gas}## and there is also an external pressure ##P_{ext}##, generally the constant atmospheric pressure. I guess we can use either pressure in the definition of work ##W##. But the pressure ##P_{gas}## can be variable: I envision a reaction where the internal pressure change with time (ex: explosion).
If the gas is free to expand/contract and change its volume ##V##, work is nonzero and there is a change in energy.
Instead of using internal energy ##E_{int}## and its change ##\Delta E_{int}##, we can clump work with the internal energy into a new state function ##H##, enthalpy, so we don't need to track work and change in system's volume: $$[\Delta E_{int} +W] = Q$$
$$[\Delta E_{int} +W] = \Delta H$$
$$\Delta H = Q$$
which means that enthalpy changes when the system absorbs or emit thermal energy ##Q## via heat (transfer or gain of thermal energy to the environment).
Does the definition of enthalpy ##H## and its changes only apply if the internal and external pressures are both equal and constant? Why? I am missing that important point...
Thanks!
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