- #36
- 23,280
- 5,682
This result is correct. But, if we reduce the terms to a common denominator, we obtain:worryingchem said:## W_f = nRT - \frac{nRTM}{M+m} ##
## W_f = nRT \frac{m}{(M+m)}##
In the calculations were have done here, we have treated Wf as the work done by the system (the gas) is the surroundings. With this sign, convention ΔU=Q-W, so thatSince it's an isothermal process dU = 0 and therefore, dq = -dw.
## Q =W_f = nRT \frac{m}{(M+m)}##
The last two items on our to do list for this version of the problem are to determine:
4. The change in Entropy of the gas between the initial and final equilibrium states
5. The heat transferred Q divided by the temperature TI at the boundary where the heat transfer takes place during the irreversible expansion process
Item 5 is trivial to do at this point, because the boundary of the system is maintained at the temperature T for then entire process, so
##\int \frac{dQ}{T_I}=\frac{Q}{T}=nR \frac{m}{(M+m)}##
Now for determining the change in entropy of the system between the same initial and final thermodynamic equilibrium states. Any ideas on how to get the change in entropy ΔS (so that we can compare it with the above integral)?
Chet