Question
One mole of an ideal monoatomic gas is taken from state A ( atm, L) to state B ( atm, L) along two different paths: (i) isothermal expansion, (ii) first isobaric expansion at atm to , then isochoric cooling at to atm.
Find the work done by the gas in each path. Are the temperatures at A and B the same?
Solution — Step by Step
Using , . At A: . At B: .
So — not isothermal! The “isothermal expansion” in path (i) is therefore a trap; we cannot directly connect A to B isothermally because they’re at different temperatures.
What the problem really intends: along path (i), the gas first cools isothermally? No — for path (i) to take A to B, it must be an isothermal and an isochoric heating, or the problem must be a curve passing through both points.
The cleanest reading: path (i) is along the curve constant only locally — but this is inconsistent with the data. The correct framing is path (i) follows ? Let’s instead solve path (ii) cleanly and note path (i) is impossible as stated.
Isobaric leg ( atm, L):
Isochoric leg ( L constant): no volume change, so .
Total work along path (ii): .
Since A and B are not on the same isotherm, no genuine isothermal process connects them. The trap in this kind of problem is to compute blindly without first verifying .
Why This Works
State variables (, , ) are fixed at A and B — but path matters for and . The check tells us instantly whether two states share an isotherm, an adiabat, or neither.
The key habit: always compute and first. It tells you which processes can or cannot connect them.
Alternative Method
For path (ii), we can also compute using (monoatomic: ), then use first law to find for each leg separately. Useful cross-check.
Common Mistake
Applying for an “isothermal” process between two states without checking that the temperatures actually agree. If , no isothermal path connects them — and the formula is meaningless. Always do the check first.