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I have a 40-amp circuit that previously was used for the oven/range in my house, though I switched to gas so the circuit is no longer in use. If I wanted to install an L6-30r (30-amp receptacle) on this circuit is it sufficient to simply swap the breaker for a 30-amp and use the existing (appears to be 8AWG) wire? Second part of question, most L6-30r's that I'm finding note that they accept 14-10AWG wire, is this the norm and if so what other options would I have if I want to connect a device to this circuit with an L6-30p?

**Edit - I should have noted this would be a dedicated circuit, no other receptacles would be installed on it. The existing receptacle for the range/oven would be removed.

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Going down to #14 is not a surprise. That is the smallest wire legal for use in any in-wall mains wiring. This would normally be 10 AWG, but there are rare edge conditions where a smaller wire circuit is permitted to be up-breakered, and the only example I can come up with is welders with a short duty cycle.

Not being able to go to #8 is also not a surprise; the terminations are not rated for aluminum wire, and the right aluminum wire for 30A is #8.

Anyway, here the answer is you get about 6 inches of #10 wire and pigtail. A red wire nut, cranked down hard, should suffice for a #8 to #10 join.

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You should be able to do what you said - the 30 amp breaker will be protecting the 30 amp receptacle and the 8 gauge cable. In order to use a receptacle that accepts only up to 10 gauge, but you have 8 gauge, you should be able to just make short, 10 gauge pigtails. These would be connected inside the receptacle box using wire nuts if appropriate size to the existing wires, and then the other end to the receptacle.

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