This problem, we're going to heat an ideal gas at constant pressure from a constant temperature
reservoir, and we want to know the total entropy change for the process. So the idea is, we
have something like a piston and cylinder. The gas starts out at 20 degrees C, we're
providing heat from a 200 degrees C reservoir, and we're going to end up with constant pressure.
So the piston moves up and we end up with the gas at 100 degrees C. So we have two entropy
changes - the entropy change of the gas and the entropy change of the reservoir. So let's
look at the gas first - ideal gas, constant pressure. So delta S for the gas is just the
heat capacity, Cp, the log of the final temperature over the initial temperature. This is true
for an ideal gas. Cp is just 7/2 R, and then of course temperatures have to be absolute.
So this is 273 plus 100; 273 plus 20. And if I substitute the numbers into this, I get
0.845 times the ideal gas constant. So that's the entropy change for the gas. Now we need the
entropy change for the reservoir. So to do that, we need to know the value of Q, and
let's determine that by doing first law energy balance. This is important, we're going to
do an energy balance on the gas because we have the heat capacity of the gas. It's a
closed system, delta U is Q plus W. We're trying to determine Q, constant pressure so
W is pressure times the volume change. Well since it's constant pressure, we can write
this as delta U, and we'll bring the PV to the other side, but I can also put the pressure
inside the delta sign because it's constant pressure. So now I've solved for Q. Delta
U plus delta PV, this is just delta H. So as we know, for a constant pressure process
in a closed system, Q is just delta H. So I can calculate the value of Q because Q is
delta H, which is Cp delta T. Cp is 7/2R, delta T - well when looking at a temperature
difference, I don't have to write it in absolute temperature because the difference is going
to be the same - 100 minus 20. So I have a value for Q, this is 280 times R. Important
thing, keep in mind this is for the gas, so let's write this Q for the gas. We want Q
for the reservoir, which is minus Q for the gas. So as we remove heat from the reservoir,
we add heat to the gas. And so this means Q for the reservoir is -280 times the gas
constant. This is now the value that we're going to substitute back into the calculation
of entropy change for the reservoir. So the reservoir's at constant temperature, that's
what we mean when we say a reservoir in thermodynamics. So delta S for the reservoir is just Q for
the reservoir over the temperature of the reservoir. Well this is -280 times R, temperature
- absolute temperature - 273 plus 200. Value of -0.592R. Delta S total, which is what the
problem asks for, is just the sum of these two values. Delta S for the gas plus delta
S for the reservoir. So I've done the calculations - used the value for the gas constant in joules
per mole kelvin. Delta S total, 2.1 joules per mole per kelvin. And it's always good
to check, it's a positive number as we expect for an irreversible process. We have a finite
temperature difference and so we expect the total entropy change, system plus surroundings,
to be greater than zero.