For a $313 g$ $"313 g"$ sample of helium gas that cools by ${41.6}^{\circ} C$ $41.6^@ "C"$ due to being compressed by $267 L$ $"267 L"$ at a constant external pressure of $1 atm$ $"1 atm"$ , if its molar heat capacity is ${20.8 J/mol}^{\circ} C$ $"20.8 J/mol"^@ "C"$ , calculate $DeltaE$ , $q$ , $w$ , and $DeltaH$ ?

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For a $313 g$ $"313 g"$ sample of helium gas that cools by ${41.6}^{\circ} C$ $41.6^@ "C"$ due to being compressed by $267 L$ $"267 L"$ at a constant external pressure of $1 atm$ $"1 atm"$ , if its molar heat capacity is ${20.8 J/mol}^{\circ} C$ $"20.8 J/mol"^@ "C"$ , calculate $DeltaE$ , $q$ , $w$ , and $DeltaH$ ?

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Answer 1 · 2015-12-24T02:17:12

Yep, this is going to be long. So bear with me! Knowing your derivations would really help here...

Well, first of all, let's get the pressure in terms of $"bar"$ , because $"1 L"cdot"bar" = "100 J"$ (you'll see what I mean later).

$P = "1.01325 bar"$

Alright, so... evidently we should know the first law of thermodynamics if we have to solve for all three of these:

$\mathbf(DeltaE = q + w)$

where $DeltaE$ is the change in internal energy caused by heat flow $q$ and work $w = -P int_(V_1)^(V_2) dV$ .

The next thing we should know is how work is defined. We know that the temperature decreased, and that it doesn't matter what $T_2$ or $T_1$ are ( $E$ is a state function).

We know that the gas was compressed, so from the perspective of the gas, work was done on the gas. That means the gas let something else work upon it. Thus, the work will turn out to be numerically positive (a negative $DeltaV$ in $-PDeltaV$ gives $w > 0$ ).

Also, I guess since the pressure is constant, using a convenient thermodynamics relation:

$DeltaH = DeltaE + Delta(PV)$

$= q + w + (PDeltaV + VDeltaP + DeltaPDeltaV)$

$= q - cancel(PDeltaV) + cancel(PDeltaV) + cancel(VDeltaP + DeltaPDeltaV)^(DeltaP = 0)$

$\mathbf(DeltaH = q_p)$

in $"J"$ .

Now what about that molar heat capacity $barC_P$ that you were given? Well, there's another nice relationship with molar enthalpy $DeltabarH$ :

$((delbarH)/(delT))_P = barC_P$

(the partial derivative of the molar enthalpy with respect to the temperature at a constant pressure is the molar heat capacity.)

Thus, at a constant pressure:

$dbarH = barC_PdT$

$\mathbf(DeltabarH = int_(T_1)^(T_2) barC_P dT)$

in $"J/mol"$ (not $"J"$ ).

Since the heat capacity will stay pretty much linear in this temperature range (we have to assume that just because we aren't given the equation... though you can probably look it up in the NIST database):

$n_"He"DeltabarH = color(blue)(q_p = n_"He"barC_PDeltaT)$

where $n_"He"$ is the $"mol"$ s of helium, and $T_2 < T_1$ . Since molar enthalpy $DeltabarH$ is in $"J/mol"$ but $DeltaE$ is in $"J"$ , we had to convert everything to $"J"$ .

Note that $\mathbf(C_P)$ is actually in $\mathbf("J/"^@"C")$ , NOT $\mathbf("J/mol"cdot""^@ "C")$ .

$\mathbf(barC_P ne C_P)$

Now we need $w$ and $DeltaE$ , which at this point we pretty much have. Take the integral of the work and that's it.

$w = -P int_(V_1)^(V_2) dV$

$color(blue)(w = -P(V_2 - V_1))$

in $"L"cdot"bar"$ , which you can convert to $"J"$ (Wikipedia the universal gas constant $R$ in $("L"cdot"bar")/("mol"cdot"K")$ and compare it in $"J"/("mol"cdot"K")$ ).

And finally:

$color(blue)(DeltaE) = q_p + w$

$= color(blue)(n_"He"barC_PDeltaT - P(V_2 - V_1))$

where $V_2 < V_1$ . You can probably tell by now that the units are, of course, $"J"$ .

Yep, that's all of it! Just use these numbers you've been given, and that should be it.

$barC_P = "20.8 J/mol"^@ "C"$
$P = "1.01325 bar"$
$DeltaT = -41.6^@ "C"$
$DeltaV = -"267 L"$
$n_"He" = "313" cancel"g He" xx "1 mol He"/("4.0026" cancel"g He") = ...$

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