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For the reversible isothermal expansion of an ideal gas: $${∆H}={∆U}=0 \tag1$$ This is obvious for the case of internal energy because $${∆U} = \frac {3}{2} n R {∆T} = 0 \tag2$$ and $${∆U} = -C_P n {∆T} = 0 \tag3$$ For the case of enthalpy it is easy to see that $${∆H} = -C_v n {∆T} = 0 \tag4$$ I've also seen $${∆H} = ∆U + ∆(PV) = ∆U + nR{∆T} = 0 \tag5$$ Now for the part I don't understand. $$dH = dU + PdV \tag6$$ $$dH = dU + nRT \frac {dV}{V} \tag7$$ $${∆H} = {∆U} + nRT \ln\frac {V_2}{V_1} \tag8$$ $${∆H} = 0 + nRT \ln\frac {V_2}{V_1} = \ln\frac {V_2}{V_1} ≠ 0\tag9$$ Clearly, it is incorrect to make the substitution $ P = nRT/V$ in going from $(6)$ to $(7)$. Why is that? I thought equation $(6)$ was always valid, and integrating such a substitution should account for any change in the variables throughout the process. Why does this not yield the same answer as $(4)$ and $(5)$?

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2 Answers 2

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$$H = U + PV \Rightarrow$$

$$dH = dU +PdV + VdP\tag{6}$$

In other words, equation 6 is missing the $VdP$ term.

$$dH = dU + nRT \frac{dV}{V} + nRT \frac{dP}{P}\tag{7}$$

$$ \Delta H = \Delta U + nRT \ln\frac{V_2}{V_1} + nRT \ln\frac{P_2}{P_1}\tag{8}$$

$$P_1 V_1 = P_2 V_2 \text{ (isothermal)}$$

$$\Delta H = \Delta U + nRT \left( \ln\frac{V_2}{V_1} + \ln\frac{V_1}{V_2} \right) = \Delta U = 0\tag{9}$$

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  • $\begingroup$ I've never seen it written with a $ VdP $ term. My book presents it as written in $ (6) $ for an ideal gas. Are there certain situations in which $ VdP $ must be included, and others where it is acceptable to just use $ PdV $? $\endgroup$
    – David
    Feb 26, 2014 at 15:21
  • $\begingroup$ Books omit the VdP term for constant pressure situations, because it is zero. All other situations require the VdP term. H = U + PV is by definition. Then, using the product rule for derivatives you immediately obtain dH = dU +PdV + VdP. $\endgroup$
    – DavePhD
    Feb 26, 2014 at 15:27
  • $\begingroup$ In the Carnot cycle, the amount of work done in moving along an isotherm is calculated without the $ VdP $ term. In looking at the PV diagram, the pressure isn't constant. How can the term be ignored in such a case then? $\endgroup$
    – David
    Feb 26, 2014 at 15:30
  • $\begingroup$ Is it because P varies as a function of V? Whereas in the situation I described it does not? The situations seem analogous to me, but they are clearly not. $\endgroup$
    – David
    Feb 26, 2014 at 15:32
  • $\begingroup$ H is enthalpy, not work. dW = Fdx = PAdx = PdV, where F is force, x is distantce, A is area. $\endgroup$
    – DavePhD
    Feb 26, 2014 at 15:41
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$$PV =nRT$$ So, for constant temperature, $dU=0$.

$H=U+PV$

The term $d(PV) = 0$ because $PV$=constant

So, $dH=0$

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