## Final 2014 - Q2B,a

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Alexis Elliott 1J
Posts: 28
Joined: Wed Sep 21, 2016 2:59 pm

### Final 2014 - Q2B,a

This question seems very similar to question we had on our midterm, I guess I am still just a bit confused as to why there is a change in entropy, when everything else (U,w,q,H) would be zero.

Since there is no change in temperature, volume and pressure, would the entropy still be 0? Or is it not 0 because we are calculating the change in entropy of the two different containers?

Thanks!

Vivian Wang 3J
Posts: 29
Joined: Wed Sep 21, 2016 2:57 pm
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### Re: Final 2014 - Q2B,a

The answer shows $\Delta S$ to not equal 0.

There is also change in volume. The changes in volume occurs for each gas.
For Argon, from 7.60L to 12.90L, and for Neon, from 5.30L to 12.90L.
Though the final volumes are the same, the initial volume of each gas is not, which must be considered when calculating $\Delta S$ for each.