# What happens when butane and methane both absorb 100 J of heat?

Two cylinders, one containing 1 mole of $\ce{C4H10}$ gas at $\pu{1 atm}$ and the other containing 1 mole of $\ce{CH4}$ gas at $\pu{1 atm}$, are at $\pu{288 K}$. If each gas absorbs $\pu{100 J}$ of heat under conditions of constant volume, which of the following is true?
(a) The temperature of the $\ce{CH4}$ increases more than the temperature of the temperature of the $\ce{C4H10}$.
(b) The internal energy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
(c) The heat capacity of the $\ce{C4H10}$ is less than the heat capacity of the $\ce{CH4}$
(d) The entropy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
(e) The heat transferred to the $\ce{C4H10}$ is greater than the heat transferred to the $\ce{CH4}$.

The correct answer is (a). I am trying to eliminate answers, but I seem to get stuck.

We know that (b) is clearly false as we've added heat and no work is done, therefore the internal energy for both gases must increase. Likewise, (d) is false as the entropy should increase as the molecules have more heat and become more disordered. Finally, (e) is incorrect as we transferred $\pu{100 J}$ of heat to both systems.

But here is where I get stuck. I can't seem to eliminate (c) from just the information given.

• Well, c) is opposite to a) so you need to understand heat capacity. Sep 8 '15 at 20:39
• @Mithoron Yes, (c) is opposite to (a), but short of looking up those values, I'm not sure how to rationalize which heat capacity is larger than the other. Sep 8 '15 at 22:43

First, consider this from a nice write-up on heat capacity:

The amount of energy needed to raise the temperature of a substance by one degree in temperature will depend upon how many different ways the atom or molecule can store the energy in the form of these thermal motions and how much energy each kind of thermal motion involves.In general, for molecules with similar bonding the bigger the molecule the more heat has to be added to raise its temperature by one degree.

Now, examine the degrees of freedom for each molecule: there are more ways to store energy in the larger butane molecule (with 36 vibrational modes in addition to the translational and rotational modes) than there are in methane (9 vibrational modes plus translational and rotational ones).

Therefore, we'd expect a greater increase in temperature for the smaller molecule than the larger one for an equivalent amount of heat added to the system.

• It may be worth mentioning that at room temperature many of these vibrational modes are "frozen", so they don't really contribute to the heat capacity. Sep 9 '15 at 13:42
• Not according to the equipartition theory and my experience modeling energy flow between vibrational modes in non-linear hydrocarbons (low-frequency bends and torsions come to mind as modes that will be populated at room temperature). While the mass difference manifests in varying translational and rotational sinks, there are plenty of vibrational modes that are populated above zero-point, not to mention Coriolis coupling (roto-vibrational mixing modes), which is certainly a non-negligible effect. Sep 9 '15 at 13:53
• Well, I never said that all of them are frozen. Sep 9 '15 at 13:55
• @IvanNeretin - I appreciate the comment, thanks! Sep 9 '15 at 14:04

To substantiate the existing qualitative answer, I made the corresponding numerical calculations using REFPROP – NIST Standard Reference Database 23, Version 9.0, which provide the following quantitative results.

$$\begin{array}{lll}\hline\text{Quantity}&\text{Unit}&\text{Methane}&&\text{Butane}\\&&\text{Initial}&\text{Final}&\text{Initial}&\text{Final}\\\hline\text{Temperature}\ T&\mathrm K&288.00&291.70&288.00&289.13\\\text{Pressure}\ p&\mathrm{atm}&1.0000&1.0129&1.0000&1.0043\\\text{Volume}\ V&\mathrm l&23.586&23.586& 22.831&22.831\\\text{Amount of substance}\ n&\mathrm{mol}&1.0000&1.0000&1.0000&1.0000\\\text{Internal energy}\ U&\mathrm{kJ}&11.847&11.947&33.150&33.250\\\text{Heat capacity}\ C_V&\mathrm{J/K}&27.050&27.178&88.677&88.944\\\hline\end{array}$$