Question

One mole of helium gas at 10 K or, one mole of helium gas at 10o
C? Figure out which one substance has a higher entropy and explain in detail why

Answer 1

Answer:

its easy

Explanation:

According to Bolztmann equation, we can relate the entropy of a system to the amount of configurations it can present, as:

S=k_B T \log Q

Where Q is the amount of microstates or configurations of the system. In the case of an ideal gas, Q is very easy to calculate. Each atom has access to a volume V, and since each particle is independent to each other the total amount of configurations is V^N, just as if I throw N dices the total amount of results (considering the order) is 6^N. Therefore, if we substitute this value of Q into the previous equation we get.

S=k_B T \log V^N= N k_B T \log V= R n T \log V

where in the last step I have divided and multiplied by the Avogadro number, so k_B turns into R and the number of particles is expressed in mol, rather than the actual number.

So now we have a problem. To obtain S I need both T and V, however there is no way I can calculate V because they don't give any information about the volume of each gas or what is its pressure. Therefore, if the gas at 10°C is at a billion atmospheres of pressure, while the one at 10K is at 0.000001 atm of pressure, then the one at 10K will have more entropy.

hope its clear

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