# Question 487bc

##### 1 Answer
Dec 28, 2015

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

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

• $\setminus m a t h b f \left(\mathrm{dH} = \mathrm{dU} + d \left(P V\right)\right)$ (1)
• $\setminus m a t h b f \left(\mathrm{dU} = \partial {q}_{\text{rev" + delw_"rev}}\right)$ (2)
• $\setminus m a t h b f \left(\partial {w}_{\text{rev}} = - P \mathrm{dV}\right)$ (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 $\setminus m a t h b f \left(q\right)$ 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 (${C}_{p} - {C}_{V} = n R$ 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 $P V = n R T$.

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

---PART A---

ENTHALPY AND INTERNAL ENERGY

In an isothermal process, we know that $\Delta T = 0$. Using (4), we get:

$\mathrm{dH} = {C}_{p} \mathrm{dT}$

$\int \mathrm{dH} = \textcolor{b l u e}{\Delta H} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{p} \mathrm{dT} = \textcolor{b l u e}{0}$

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

$\mathrm{dU} = {C}_{V} \mathrm{dT}$

$\int \mathrm{dU} = \textcolor{b l u e}{\Delta U} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{V} \mathrm{dT} = \textcolor{b l u e}{0}$

...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 $\Delta U = - \Delta \left(P V\right)$, so, using (2) with (1):

$\partial {q}_{\text{rev" + delw_"rev}} = - \left(P \mathrm{dV} + V \mathrm{dP}\right)$

and then using (3):

$\partial {q}_{\text{rev}} - \cancel{P \mathrm{dV}} = - \cancel{P \mathrm{dV}} - V \mathrm{dP}$

$\partial {q}_{\text{rev}} = - V \mathrm{dP}$

$\int \partial {q}_{\text{rev}} = - {\int}_{{P}_{1}}^{{P}_{2}} V \mathrm{dP}$

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, $V = \frac{n R T}{P}$, where $n$, $R$, and $T$ are all constant:

$\textcolor{b l u e}{{q}_{\text{rev}}} = - {\int}_{{P}_{1}}^{{P}_{2}} \frac{n R T}{P} \mathrm{dP}$

$= - n R T {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

$\textcolor{b l u e}{= - n R T \ln | \frac{{P}_{2}}{{P}_{1}} |}$

REVERSIBLE WORK

For reversible work, note that:

$\mathrm{dU} = \partial {q}_{\text{rev" + delw_"rev}}$

${\cancel{\Delta U}}^{0} = {q}_{\text{rev" + w_"rev}}$

thus:

$\textcolor{b l u e}{{w}_{\text{rev" = -q_"rev}}}$

---PART B---

REVERSIBLE HEAT FLOW, AND INTERNAL ENERGY

In an adiabatic process, we should know that $\textcolor{b l u e}{\partial {q}_{\text{rev}} = 0}$. Thus:

$\mathrm{dU} = {C}_{V} \mathrm{dT} = \partial {w}_{\text{rev}} = - P \mathrm{dV}$

In this case, ${T}_{1} = \text{300 K}$ and ${T}_{2} = \text{102 K}$. 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 $\mathrm{dU} = {C}_{V} \mathrm{dT}$:

$\textcolor{b l u e}{\Delta U} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{V} \mathrm{dT}$

$= \textcolor{b l u e}{\frac{3}{2} R \left({T}_{2} - {T}_{1}\right)}$

REVERSIBLE WORK

Next, we can use the relationship recently established to determine ${w}_{\text{rev}}$. Since $\partial {q}_{\text{rev}} = 0$, $\partial w = \mathrm{dw} = \mathrm{dU}$ (work becomes an exact differential), and:

$\textcolor{b l u e}{{w}_{\text{rev}} = \Delta U}$

ENTHALPY

Finally, we still need $\Delta H$! Note that again, the enthalpy depends only upon the temperature. Not the pressure, nor the volume. Thus, we can use the definition $\mathrm{dH} = {C}_{p} \mathrm{dT}$:

$\Delta H = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{p} \mathrm{dT}$

Thus:

$\textcolor{b l u e}{\Delta H} = {C}_{p} \left({T}_{2} - {T}_{1}\right)$

$= \left({C}_{V} + n R\right) \left({T}_{2} - {T}_{1}\right)$

$= \left(\frac{3}{2} R + R\right) \left({T}_{2} - {T}_{1}\right)$

$= \textcolor{b l u e}{\frac{5}{2} R \left({T}_{2} - {T}_{1}\right)}$