## When C, Heat Capacity is not given?

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

VivianaHF2L
Posts: 29
Joined: Fri Sep 28, 2018 12:20 am

### When C, Heat Capacity is not given?

For problem 9.3 the formula for change in temperature is needed but how are we suppose to find Heat Capacity when the number of moles are not given? Without the moles the formula for solving C is not possible.

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

### Re: When C, Heat Capacity is not given?

If you are referencing problem 9.3 in 6th Edition textbook, for part (a), you would use the equation deltaS=deltaQ/T. You would get deltaQ = +65J and T = 25 + 273 = 298 K. This gives you a deltaS of 65/298 = 0.22J/K

You would use the same process for part (b), except with 373 K instead of 298.

For part (c), the entropy change in part (a) was higher, because temperature is in the denominator, and thus a lower temperature will result in a higher change in entropy.