Question

Question #487bc

Answer 1

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))`