I Question regarding dh, du, cp, cv for ideal gases

[Uchida]

Hi,

Considering the question bellow from a government work selection process:Check the FALSE alternative on the use of thermodynamic properties.

1. In a cylinder-piston type system, the variation of the enthalpy property (Δh) is usually applied to determine the heat (per kilogram) exchanged with the neighborhood (q = Δh), but only for processes under constant pressure.
2. The product of specific heat at constant volume (Cv), by varying the temperature of a system (ΔT), is used to determine the variation of specific internal energy (Δu = Cv ΔT), but only when the volume is kept constant (isovolumetric process ) and when there is no phase change.
3. For the ideal gas model, it is usual for the specific heat property at constant pressure Cp to be considered independent of the fluid pressure.
4. For the ideal gas model model, the specific heat property at constant volume Cv is considered to be independent of the fluid pressure.

I consider both options 1 and 2 to be false, because, for ideal gases, cv and cp are only dependent of T, therefore, given that

dh = cp.dT
du = cv.dT

dh and du expressed as above are true whether or not the process is isobaric / isochoric.

Also, continuing with 1º Law of Thermodynamics:

𝛿q = du + 𝛿w = cp.dT = cv.dT + p.dv = dh

For a process with frontier work, p.dv is not necessarily isobaric. Nervertheless, dh = 𝛿q would still be valid, correct?

Therefore, in my view, both the statements of alternatives '1' and '2' are false.

However, in the answer to the question, only '2' is considered to be false.Where's my mistake?

Thank you.

 

[Chestermiller]

Your mistake is assuming that, for a non-isobaric process, $q=\Delta h$. $\Delta h=C_p\Delta p$ for an ideal gas undergoing an arbitrary process, but not q.

 

[vanhees71]

In the canonical ensemble (particle number strictly fixed) you have two independent thermodynamical quantities, and the potentials for the various processes are all derived from
$$\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V.$$
That means that U is the right potential for isochoric processes (processes at constant volume), i.e., for $\mathrm{d} V=0$ you have $\mathrm{d} U=T \mathrm{d} S=\mathrm{d} Q$. So in this case you get the specific heat at constant volume,
$$C_V=\left (\frac{\partial U}{\partial T} \right)_V.$$
Since for an ideal gas
$$U=\frac{f}{2} N k T,$$
with $f=3$ for monatomic, $f=5$ for two-atomic and $f=6$ for $n$-atomic ($n \geq 3$) gases, you find
$$C_V=\frac{f}{2} N k T.$$
From this you can also derive the entropy as a function of $V$ and $T$ relative to a state $V_0$ and $T_0$, because you have
$$C_V=T \left (\frac{\partial S}{\partial T} \right)_V=\text{const} ; \Rightarrow \; S-S_0=C_V \ln \left (\frac{T}{T_0} \right) + \tilde{S}(V).$$
Further for an isothermal process you have (using also the equation of state $p V=N k T$)
$$\mathrm{d} U=0=T \mathrm{d} S-p \mathrm{d} V \; \Rightarrow \; \left (\frac{\partial S}{\partial V} \right)_T = \frac{p}{T} = \frac{N k}{V} \; \Rightarrow \; \tilde{S}=N k \ln \left (\frac{V}{V_0} \right).$$
Thus we have
$$S=S_0 + C_V \ln \left (\frac{T}{T_0} \right) + N k \ln \left (\frac{V}{V_0} \right).$$
For isobaric changes (i.e., changes at constant pressure) you need the enthalpy
$$H=U+p V \; \Rightarrow \; \mathrm{d} H=T \mathrm{d} S+ V \mathrm{d} p.$$
From this you get for $\mathrm{d} p=0$
$$\mathrm{d} H=\mathrm{d} Q = T \left (\frac{\partial S}{\partial T} \right)_p.$$
With the equation of state we get from the above expression for $S$
$$S=S_0 + C_V \ln \left (\frac{T}{T_0} \right) + N k \ln \left (\frac{T p_0}{T_0 p} \right) = S_0 + (C_V+N k) \ln \left (\frac{T}{T_0} \right) - N k \ln \left (\frac{p}{p_0} \right).$$
From this you find
$$C_P= T \left (\frac{\partial S}{ \partial T} \right)_P=C_V+N k= \left ( \frac{f}{2}+1 \right) N k.$$

 

[Uchida]

Thanks for your answers.

I see the mistake I´ 've made.

Starting from H = U + PV:

h = u + pv

Then

dh = du + p.dv + v.dp = 𝛿q + v.dp

so, dh = 𝛿q only if the process is isobaric, where dp = 0 from the v.dp term.