I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
They essentially mean:
These are generally used as assumptions to simplify an experiment, or simply to put a situation into a more manageable problem in a chemistry course. Specifically, you should see something like this in a Physics course or a Physical Chemistry course (among others, if applicable).
I’ve written out some examples of what you can figure out by knowing that the situation is assumed to be such that these terms hold true.
First, some definitions:
And the following are some equations we’ll end up using (Physical Chemistry: A Molecular Approach, McQuarrie).
Enthalpy vs. Internal Energy:
##mathbf(DeltaH = DeltaU + Delta(PV))##
First Law of Thermodynamics:
##mathbf(DeltaU = q + w)##
ISOTHERMAL PROCESS
Here, ##DeltaT = 0##.
For an ideal gas, that automatically means the change in internal energy ##color(blue)(DeltaU = 0)## and the change in enthalpy ##color(blue)(DeltaH = 0)##, because for an ideal gas, the internal energy and enthalpy are only dependent on the temperature.
By using this equation, you can determine the relationship between heat flow and work now, and it greatly simplifies the problem:
##cancel(DeltaH) = cancel(DeltaU) + Delta(PV)##
##= Delta(PV)##
##= PDeltaV + VDeltaP##
##color(blue)(w_”rev” = -PDeltaV = VDeltaP)##
Furthermore, using ##DeltaU = q + w##, the first law of thermodynamics:
##0 = q_”rev” + w_”rev”##
##color(blue)(q_”rev” = -w_”rev” = -VDeltaP)##
You couldn’t say these were true unless it was an ideal gas at isothermal conditions! In my opinion, this is one situation that I find fairly straightforward.
ADIABATIC PROCESS
Here, ##color(blue)(q = 0)##, so using the first law of thermodynamics again:
##color(blue)(DeltaU) = cancel(q_”rev”)^(0) + color(blue)(w_”rev” = -PDeltaV)##
Additionally, using the enthalpy equation from earlier:
##DeltaH = DeltaU + Delta(PV)##
##= w_”rev” + PDeltaV + VDeltaP##
##= -cancel(PDeltaV) + cancel(PDeltaV) + VDeltaP##
Thus, ##color(blue)(DeltaH = VDeltaP)## when a process upon an ideal gas is adiabatic.
ISOVOLUMETRIC PROCESS
A similar situation arises when ##DeltaV = 0##, because it means expansion/compression work ##color(blue)(w_”rev” = 0)## (see the above usages of ##w = -PDeltaV##?):
##color(blue)(DeltaU = q_V)##
where ##q_V## is (presumably reversible) heat flow at a constant volume.
However, it does not matter for enthalpy because if you recall from the adiabatic process, let us work backwards from the relation ##DeltaH = VDeltaP##. If it was not adiabatic, ##q ne 0##, thus:
##color(blue)(DeltaH = q_V + VDeltaP)##
which does not depend on ##DeltaV##.
This can be a fairly challenging situation, because you don’t have any easy relationship where you can just do a simple integration of a ##dT## term, or a ##dV## term, or similar (giving e.g. ##PDeltaV##, ##VDeltaP##, etc).
Unless you know the following relationships, it might be difficult to figure this out in full.
##color(blue)(DeltaH) = int_(T_1)^(T_2) C_pdT##
##= color(blue)(C_p(T_2 – T_1))##
where ##C_p## is the constant-pressure heat capacity (doesn’t have to be use in a constant-pressure situation though). For an ideal gas, it is assumed to be a constant across small temperature ranges.
From determining ##DeltaH##, you can fairly easily determine ##q##, and thus ##DeltaU##. Conveniently, you also have this relationship:
##color(blue)(DeltaU) = int_(T_1)^(T_2) C_VdT##
##= color(blue)(C_V(T_2 – T_1))##
where ##C_V##, the constant-volume heat capacity (doesn’t have to be use in a constant-volume situation though), is based on the degrees of freedom for an ideal gas. Without going too much into this, ##C_V = 3/2 R## for a monatomic ideal gas, where ##R## is the universal gas constant, and ##C_p = C_V + nR##.
CHALLENGE: What do you imagine will be the relationships for ##DeltaU##, ##DeltaH##, ##q_”rev”##, and ##w_”rev”## for an isobaric process? Hint: It means constant pressure, and it is similar to the isovolumetric process.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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