# Question #487bc

##### 1 Answer

*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 **(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 anideal gasdepends upon"only the temperature(McQuarrie, Ch. 19-4).In other words, when the temperature is constant, enthalpy and internal energy areboth#0# for anideal gas.

**REVERSIBLE HEAT FLOW**

Now, solving for the reversible heat flow, using **(1)**, that means **(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 ** 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

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

In this case,

#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 **exact differential**), and:

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

**ENTHALPY**

Finally, we still need

#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))#