Question #487bc

Disclaimer: Yes, this will be long! No getting around it!

It really helps to know the derivations. So, what do we know?

• #\mathbf(dH = dU + d(PV))# (1)
• #\mathbf(dU = delq_"rev" + delw_"rev")# (2)
• #\mathbf(delw_"rev" = -PdV)# (3)
• #\mathbf(((delH)/(delT))_P = C_P# (4)

• #\mathbf(((delU)/(delT))_V = C_V# (5)

• An isothermal situation assumes a constant temperature during the expansion process.

• An adiabatic situation assumes no heat flow #\mathbf(q)# contributes to the internal energy #U#.

• The volume might have changed, but we don't know how, exactly. Even though we were given #C_V#, we can't assume that it is a constant-volume situation since we can convert from #C_V# to #C_p# pretty easily (#C_p - C_V = nR# for an ideal gas).

• The pressure decreased, i.e. it is NOT constant. That means that we should expect to eventually somehow use the relationship #PV = nRT#.

So, having said that, let's see...

---PART A---

ENTHALPY AND INTERNAL ENERGY

In an isothermal process, we know that #DeltaT = 0#. Using (4), we get:

#dH = C_pdT#

#int dH = color(blue)(DeltaH) = int_(T_1)^(T_2) C_pdT = color(blue)(0)#

...and using (5), we get:

#dU = C_VdT#

#int dU = color(blue)(DeltaU) = int_(T_1)^(T_2) C_VdT = color(blue)(0)#

...for an ideal gas. REMEMBER THIS:

"The energy of an ideal gas depends upon only the temperature" (McQuarrie, Ch. 19-4). In other words, when the temperature is constant, enthalpy and internal energy are both #0# for an ideal gas.

REVERSIBLE HEAT FLOW

Now, solving for the reversible heat flow, using (1), that means #DeltaU = -Delta(PV)#, so, using (2) with (1):

#delq_"rev" + delw_"rev" = -(PdV + VdP)#

and then using (3):

#delq_"rev" - cancel(PdV) = - cancel(PdV) - VdP#

#delq_"rev" = -VdP#

#intdelq_"rev" = -int_(P_1)^(P_2) VdP#

We don't know how the volume changed, just how the pressure changed, so we have to substitute #V# for something else using the ideal gas law. So, #V = (nRT)/P#, where #n#, #R#, and #T# are all constant:

#color(blue)(q_"rev") = -int_(P_1)^(P_2) (nRT)/P dP#

#= -nRT int_(P_1)^(P_2) 1/P dP#

#color(blue)(= -nRT ln|(P_2)/(P_1)|)#

REVERSIBLE WORK

For reversible work, note that:

#dU = delq_"rev" + delw_"rev"#

#cancel(DeltaU)^(0) = q_"rev" + w_"rev"#

thus:

#color(blue)(w_"rev" = -q_"rev")#

---PART B---

REVERSIBLE HEAT FLOW, AND INTERNAL ENERGY

In an adiabatic process, we should know that #color(blue)(delq_"rev" = 0)#. Thus:

#dU = C_VdT = delw_"rev" = -PdV#

In this case, #T_1 = "300 K"# and #T_2 = "102 K"#. Furthermore, we again need to realize that the internal energy of an ideal gas depends only upon the temperature. Not the pressure, nor the volume. Therefore, we can use the definition #dU = C_VdT#:

#color(blue)(DeltaU) = int_(T_1)^(T_2) C_VdT#

#= color(blue)(3/2 R (T_2 - T_1))#

REVERSIBLE WORK

Next, we can use the relationship recently established to determine #w_"rev"#. Since #delq_"rev" = 0#, #delw = dw = dU# (work becomes an exact differential), and:

#color(blue)(w_"rev" = DeltaU)#

ENTHALPY

Finally, we still need #DeltaH#! Note that again, the enthalpy depends only upon the temperature. Not the pressure, nor the volume. Thus, we can use the definition #dH = C_pdT#:

#DeltaH = int_(T_1)^(T_2) C_pdT#

Thus:

#color(blue)(DeltaH) = C_p(T_2 - T_1)#

#= (C_V + nR)(T_2 - T_1)#

#= (3/2 R + R)(T_2 - T_1)#

#= color(blue)(5/2 R(T_2 - T_1))#