Gain is absolutely the ONLY important part of PSRR. Essentially what you are saying is how much can an op-amp when feeding back a signal cancel out any ripples introduced from the power supply, not from the input of the circuit..
Lets take a simple example: an ideal (infinite open loop gain) voltage follower (output tied directly to the inverting input, fed from the non inverting input). The circuit has a closed loop gain of 1, but the feedback (since the overall gain is SOOO high) will mean that any power supply ripple will be canceled due to the feedback forcing the non inverting and inverting inputs to be in perfect lockstep..
But take the SAME example, but make the OPEN loop gain of the opamp 1, still with closed loop gain of 1, then suddenly the op amp can't keep up with the changes between the non-inverting input and the output-inverting input. And hence all ripple from the power supply would be visible on the output (essentially the op-amp would turn into a noise source with the noise being the coupled power supply ripple)
I understand HOW stevenvh could say that the gain is not meaningful, because he meant CLOSED loop gain... But the gain of question is OPEN loop gain, and YES, that is EVERYTHING in PSRR.
EDIT: And to answer your question, just to follow up slightly here, the PSRR is related to open loop gain, but the more closed loop gain you introduce, the more power supply ripple you will get on the output (hence the 60dB you reference above)
Here is why: Same example I give above, except this time you have a REAL op amp, (finite open loop gain), and resistors in your feedback path, meaning you have a closed loop gain of some value, say 6dB. Since the resistors behave as a voltage divider, the op-amp has to OVERCOMPENSATE for the power supply ripple being fed back to the non-inverting input. If it can only compensate for 100dB of power supply ripple, you will only get 94dB of rejection. The more closed loop gain you introduce, the less of the power supply ripple you are able to reject.
The whole conversation stems from the separate meanings of open loop and closed loop gain.
2nd EDIT: And the way that you get 60dB, or I get my 94dB is that you have to realize you have to convert dB BACK so for example you need to use
$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{6}{20}}\right) = 94 \mathrm{dB}
$$
$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{40}{20}}\right) = 60 \mathrm{dB}.
$$
And YES the other guy who said it should be 1mV not 1µV on Wikipedia is correct.
You cannot just connect the probe across inputs to the inamp and expect good results.
Operational amplifiers need a path to ground from each input. So the second circuit is better.
the 741 is totally unsuitable, but the LMC6041 in the notes should work well for this circuit. (2 fA is typical rating though, use the guaranteed value of 4 pA for design unless you want your circuit to only mostly work... 4pA through 300Mohm sensor is 1.6mV offset.)
LMC6041 will be happy with a single supply, so if the -5V supply is inconvenient have a look at this circuit: What is the purpose of this op amp?
I suggest you stop simulating and actually build the circuit. LMC6041 is $2 from a distributor or ask National, they'll give free samples for school projects. You don't need to hook up the ADC yet, just get the probe and signal conditioning amplifier working. Read the output with a voltmeter. Just like software, breaking the problem into small pieces is the way to make progress.
Best Answer
Question 1:
Yes the method sounds correct. But, you do not need to add a new VDC source in series with the fixed one. Just set AC magnitude of your 1.8V DC source to 1.
Question 2:
The PSRR+ of such opamps will be bad because looking from the supply to the output, it looks like a common base amplifier! Improving PSRR is a very complicated but interesting topic, but most of it boils down to trading off headroom for better PSRR. For a start, look into other opamp topologies. Generally speaking, cascoded outputs offer better PSRR etc. If you have headroom, you can basically precede your opamp with a NMOS regulator. This will essentially cascode your whole opamp from supply noise. Ofcourse, headroom might be an issue now.
Another more sophisticated way to improve it for your opamp, is to some how move the gate of PM67(and/or any other transistor that looks like a common base amplifier from the supply) in sync with the noise at its Source, this way: $$\Delta Vgs = 0$$ To do this you can try a small capacitor between source and gate or even create a secondary gain loop that injects correlated supply noise at the gate of PM67. Yet another way is to drive the body of PM67 or NM28 in a way that cancels supply noise