expansion of ideal gas

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

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

expansion of ideal gas

When calculating the total entropy change for the expansion of an ideal gas, why can we assume that deltaU=0 and therefore q=-w? I don't understand the logic behind this and may be missing something.

Sara Flynn 2C
Posts: 60
Joined: Fri Sep 28, 2018 12:23 am

Re: expansion of ideal gas

if an ideal gas is expanding isothermally that means that it is happening at a constant temperature and this can happen because as the system is expanding and the volume is increasing and the work is being done, the temperature is being kept constant by heat that is flowing into the system to maintain the temperature. So the work is equal to the amount of heat flowing back into the system