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Note: This is not a homework problem; I am not asking how to solve this problem; and not even why my solution is wrong, in a sense. Just ignore the numbers in the "background" part. I included the question from the textbook as a support of the claim that "at constant volume work done is zero. " I am asking this from a conceptual point of view. I will continue this at the end of the post...

Background: I was solving this question: Calculate the final temperature and the change in internal energy when $500 . \mathrm{J}$ of energy is transferred as heat to $0.900 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})$ at $298 \mathrm{~K}$ and $1.00 \mathrm{~atm}$ at $(\mathrm{a})$ constant volume (b) constant pressure. Treat the gas as ideal.

The solution(1)(2) to the first part given in the book is as follows:

From $C_{V, \mathrm{~m}}=\frac{5}{2} R$ and $C_{P, \mathrm{~m}}=C_{V, \mathrm{~m}}+R$

$C_{V, \mathrm{~m}}=\frac{5}{2}\left(8.3145 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\right)=20.79 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}$ $C_{P, \mathrm{~m}}=\frac{7}{2}\left(8.3145 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\right)=29.10 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}$

(a) From $\Delta T=q / n C_{V, \mathrm{~m}}$, $$ \Delta T=\frac{500 \cdot \mathrm{J}}{(0.900 \mathrm{~mol}) \times\left(20.79 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\right)}=+26.7 \mathrm{~K} $$ Hence the final temperature $$ T=298+26.7 \mathrm{~K}=325 \mathrm{~K}, \text { or } 52^{\circ} \mathrm{C} $$ From $\Delta U=q$ at constant volume, $\Delta U=+500 \mathrm{~J}$

Question: I have only a problem in the first part; in the last line where the author says that "From $\Delta U=q$ at constant volume", I don't understand how this could be.

Clarification: We know that $H = U + PV$, now as I am not told anything about pressure in the first case I will replace $PV$ by $nRT$. Therefore I get: $$\Delta H =\Delta U + \Delta (nRT).$$

Now since moles and gas constant are constant, I get: $$\Rightarrow \Delta H =\Delta U + nR \Delta T$$

$\Delta T=26.7 \mathrm{~k}, n=0.900 \mathrm{~mol}, R=8.314\mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}$ and $\Delta H= 500 \mathrm{~J}$ we get: $$\Delta U=500-8.314\cdot26.7=278.0162$$

Clearly $\Delta H \not= \Delta U$. How could this be?

Note continued: The conceptual formula is how could enthalpy change be equal to internal energy change at constant volume; surely it can't from my method of solution; there are no errors whether relating to mathematics or physics. From another point of view, it is also clear that there is no work done since there is no volume expansion as the volume is constant, but from others, there is some work. Where am I wrong? Again I am not asking where my mistake is in homework but concept.

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  • $\begingroup$ For constant volume transformations, the relevant thermodynamic potential is $U$, not $H$. You have to work directly with $U$, without bothering about $H$. $\endgroup$ Commented Jul 18, 2021 at 9:42
  • $\begingroup$ @GiorgioP I don't quite understand you, please elaborate on your comment. Thanks. $\endgroup$
    – Osmium
    Commented Jul 18, 2021 at 9:47
  • $\begingroup$ Before, you should recast your question in a form more appropriate for this site. In its present form, it looks too much like a homework-like question, with a high probability of been closed. $\endgroup$ Commented Jul 18, 2021 at 9:52
  • $\begingroup$ @GiorgioP isn't entropy defined for every system? Do you mean to say that the formula of entropy is not valid for constant volume cases? If it is valid for all systems, then why is my solution not correct. If it is not valid....well how could that be? $\endgroup$
    – Osmium
    Commented Jul 18, 2021 at 9:55
  • $\begingroup$ @GiorgioP does it still look like a homework question? $\endgroup$
    – Osmium
    Commented Jul 18, 2021 at 10:15

2 Answers 2

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The problem statement defines $q = 500$ J, $n_{O_2} = 0.900$ mol, $T_o = 298$ K, and $p_{o,O_2} = 1.00$ atm. The gas is to be taken as ideal.

These definitions apply (IUPAC convention).

$$ \Delta U = q + w $$ $$ \Delta H = \Delta U + \Delta (pV) $$

For a REVERSIBLE process, mechanical work is below with $p_{int}$ as the system pressure.

$$ w_{mech} = - \int\ p_{int} dV $$

Finally, for the case of an ideal gas with constant (molar) specific heat capacity

$$ \bar{C}_p = \bar{C}_V + R$$

$$ \Delta U \equiv n\bar{C}_V\Delta T \ \ \ \ \Delta H \equiv n\bar{C}_p\Delta T $$

The first expression is always true of an ideal gas. The latter two expressions are always true for an ideal gas with a constant specific heat capacity regardless of whether the path taken is reversible or irreversible.

Part a is at constant volume, meaning $\Delta V = 0$ and $w_{mech} = 0$. Apply the first law to find

$$\Delta U = q_V = n\bar{C}_V\Delta T$$

where the subscript $_V$ helps clarify that the heat flow is at constant volume. Additionally

$$\Delta_V H = q_V + \Delta (pV) = q_V + nR\Delta T = n\bar{C}_p\Delta T $$

The second term on the far right in the second step is obtained by expansion of $\Delta (pV)$ for the ideal gas law in a closed system.

$\Rightarrow$ all unknown values can be determined

Part b is at constant pressure, meaning $q = q_p$. The expansion for enthalpy from internal energy gives

$$\Delta_p H = \Delta U + \Delta (pV) = q_p + w_{mech} + \Delta (pV) $$

Pressure is constant. Take a reversible path (since enthalpy is a state function) to obtain

$$\Delta_p H = q_p - p\Delta V + p \Delta V = q_p = n\bar{C}_p\Delta T $$

Given $\Delta T$, we find

$$\Delta_p U = q_p - p\Delta V \equiv n\bar{C}_V\Delta T$$

$\Rightarrow$ all unknown values can be determined

Clarifications to avoid typical conceptual mistakes:

  • $\Delta U \equiv n\bar{C}_V\Delta T$ and $\Delta H \equiv n\bar{C}_p\Delta T$ ONLY for ideal gases with constant specific heat capacity

  • $w_{mech} = -\int\ p_{int} dV$ ONLY for reversible processes

  • $\Delta (pV) = nR\Delta T$ ONLY for ideal gases in closed systems with no chemical reactions

  • $q = q_V$ in one process and $q = q_p$ in another process NEVER means $\Delta U = \Delta H$ in either process entirely by itself

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  • $\begingroup$ Your answer is excellent but I don't understand what you mean by the first clarification point. Do you mean that molar specific heat are defined only for ideal gases? Thanks. $\endgroup$
    – Osmium
    Commented Jul 18, 2021 at 14:21
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    $\begingroup$ The values of molar specific heat capacity for any substance depend on temperature and pressure. For an ideal gas, molar specific heat capacities depend only on temperature. A further assumption of constant molar specific heat capacity gives the simplest defining relationships for internal energy and enthalpy. You MUST have both assumptions. Just saying that specific heat capacity is constant still means for example that internal energy depends on both temperature and volume (e.g. for a non-ideal gas). $\endgroup$ Commented Jul 18, 2021 at 14:58
  • $\begingroup$ Regarding Berger dot 3, what if a chemical reaction occurs in which n changes at constant temperature? $\endgroup$ Commented Jul 18, 2021 at 18:02
  • $\begingroup$ @ChetMiller fixed thanks $\endgroup$ Commented Jul 18, 2021 at 19:16
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The first law of thermodynamics for an ideal gas tells us that $$\Delta U=nC_v\Delta T=q-w=q-\int{pdV}$$For constant volume, this reduces to $$\Delta U=nC_v\Delta T=q=500\ J$$and, as you found, $$\Delta T=26.7\ C$$ Now for $\Delta H$:$$\Delta H=\Delta U+nR\Delta T=500+(0.9)(8.314)(26.7)=700\ J$$Alternately, for an ideal gas, $$\Delta H=nC_p\Delta T=(0.9)(29.1)(26.7)=700\ J$$ So, clearly, $\Delta H\neq\Delta U$ for this process. So who's wrong now?

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  • $\begingroup$ But that doesn't answer where the mistake is my procedure. Isn't the definition of enthalpy as I mentioned? Isn't the change in enthalpy for the case correct, as I wrote it? Isn't it that one can do a question in many ways, but as science is consistent the person will get the same answer, provided the method of doing is correct? $\endgroup$
    – Osmium
    Commented Jul 18, 2021 at 13:12
  • $\begingroup$ You seem to have set $\Delta H$ equal to q, which, for this case it is not. What could possibly have made you think that it was? $\endgroup$ Commented Jul 18, 2021 at 13:23
  • $\begingroup$ And your point is?? $\endgroup$ Commented Jul 18, 2021 at 18:03
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    $\begingroup$ You asked why I thought $q=\Delta H$ for constant volume case, so my reply was intended to mean that "Yes! I did a mistake, indeed a very silly mistake. Thanks for answering." $\endgroup$
    – Osmium
    Commented Jul 19, 2021 at 0:55

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