Sample Quiz #3
A rigid tank contains an ideal gas at 50oC. The paddle wheel (shown) does 200 kJ of
work on the gas. The temperature of the gas remains constant during this process as a
result of heat transfer between the system and the surroundings at 25oC. Find the amount
of entropy generated during the process.
Ideal Gas
50oC
Q
Given the statement of the problem, we want S gen = dS sys + dS surr
We begin by considering the system. This is a closed system process which
involves an ideal gas. Therefore:
dS sys = s 2 − s1 = C v ln
T2
v
+ R ln 2
T1
v1
Since the process is both isothermal and isometric, dSsys = 0.
Now we consider the surroundings. The surrounds undergo an isothermal heat
transfer process. Therefore:
dS surr =
Qsurr
Tsurr
We find the heat transfer to the surroundings from the first law for a closed
system process:
∂Q = dE + ∂W
where dE = 0 because the PE and KE are negligible and internal energy is only a
function of temperature (and temperature doesn’t change in this process). Thus:
∂Q = ∂W = −200kJ since work is done on the system (so heat should be leaving
the system – hence the negative sign).
So S gen = dS surr =
Qsurr 200kJ
=
= 0.671kJ / kg
Tsurr
298K
Note the heat transfer according to the surroundings is positive.