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One mole of benzene at $\pu{20^\circ C}$ is adiabatically mixed with 2 moles of toluene at $\pu{40^\circ C}$. The heat capacity of benzene is $\pu{84 J K-1 mol-1}$ and the heat capacity of toluene is $\pu{96 J K-1 mol-1}$. If benzene and toluene form an ideal solution, compute the entropy change for the process.

I tried to use

$$\Delta S = n C \ln{(T_\mathrm f/T_\mathrm i)}$$

where $C$ is the heat capacity and $T_\mathrm f$ and $T_\mathrm i$ are the final and initial temperatures of the liquids, respectively. However, I am unsure of what the final equilibrium temperature of the mixture comes to.

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  • $\begingroup$ In addition to the temperatures of the liquids equilibrating (which results in an entropy change), you also need to include the entropy change due to mixing. $\endgroup$ – Chet Miller Oct 15 '17 at 12:03
  • $\begingroup$ So what do you get for the entropy of mixing, and how does that compare with the entropy change due to heat transfer? $\endgroup$ – Chet Miller Oct 15 '17 at 14:05
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Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.

The absolute amount of heat then released by toluene equals to the absolute amount of heat gained by benzene:

$|Q_\mathrm t| = |Q_\mathrm b|$

Considering signs, it will look like this:

$Q_\mathrm t = -Q_\mathrm b$

By definition, $Q = nc\Delta T$. Thus,

$Q_\mathrm t = 2\cdot96\cdot(40-T)$, and

$Q_\mathrm b = 1\cdot84\cdot(20-T)$

Plug in the latter two equations into the second one, and here's what you get:

$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$

Now just solve it for $T$. And plug that in your equation as $T_\mathrm f$.

Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.

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    $\begingroup$ This does not provide the complete solution to the problem. The entropy of mixing also needs to be considered. $\endgroup$ – Chet Miller Oct 15 '17 at 12:05
  • $\begingroup$ @ChesterMiller It's homework so not a problem usually. $\endgroup$ – Mithoron Oct 15 '17 at 15:56
  • $\begingroup$ The entropy of mixing still needs to be included. $\endgroup$ – Chet Miller Oct 15 '17 at 17:13
  • $\begingroup$ @ChesterMiller how might one calculate the entropy of mixing, then? $\endgroup$ – Jonathan Smith Oct 17 '17 at 14:52
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    $\begingroup$ @JonathanSmith well that's easily googleable en.wikipedia.org/wiki/Entropy_of_mixing $\endgroup$ – wolphram Oct 17 '17 at 15:01

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