## one thermodynamic problem

kolndom
Posts: 19
Joined: Tue Oct 25, 2005 11:29 am

### one thermodynamic problem

I find it hard to analyze the system (1)+(2), the system (3)+(4), and the system (6).

*Th8,9 Question 8-9 refer to the following processes involving systems labeled by numbers (1) through (8).
A bar of iron (1) at 300K is brought into thermal contact with a body (2) at 400K, the two being thermally isolated from all other systems.
An ideal gas (3) is compressed reversibly while in contact with a reservoir (4), the two being thermally isolated from all other systems.
A body of water (5) freezes reversibly.
A container of water (6) is stirred and its temperature increases by 1K.
A chemical reaction takes place in an isolated system (7).
A Carnot engine (8) operates in a cycle.
8. For which of the following systems does the entropy decrease?
(A) 1
(B) 4
(C) 5
(D) 6
(E) 7
9. For which of the following systems does the entropy increase?
(A) 2
(B) 3
(C) 8
(D) 1 and 2 combined
(E) 3 and 4 combined

yosofun
Posts: 87
Joined: Mon Oct 31, 2005 5:50 am
for systems 1 and 2, you can calculate the entropy (for each object)
from
dS = dQ/T
since Q = m c dT, once u integrate it, u get mc ln(T_2/T_1)
assuming that the bodies are the same mass, you can calculate the
final temperature T_2 from averaging the two initial temperatures ..
(T_1 is the initial temperature of each object)

for sys 3 and 4, you have an ideal gas expanding in an isothermal
process. this means that the internal energy is zero. (to wit: u = mc
delta T for ideal gas) thus, by the first law, you have dq = dw) work
for isothermal process is just P_1V_1 ln (V_2/V_1) ... there are
different forms to this formula if you recall that because the
process is isothermal, by the ideal gas equation, u have nRT_1 =
nRT_2 = P_1V_1 = P_2V_2

(so.... once u have dq ... plug it into the dS equation above)

PS: P_1 stands for P subscript 1 ...

ok hopefully this is clear.