[orgo814]

One mole of an ideal gas at STP is heated at constant pressure to three times its original temperature, and then adiabatically expanded until it reaches its original temperature. What volume does it now occupy?

A bit confused on this. My only method I could think of would be to find the temperature at 3x original temperature (273 x 3) and then use delta U = (n) (Cv,m)(deltaT) assuming heat capacity is 3/2R. Then, since delta U = W because adiabatic calculating the volume change after calculating the initial volume at STP (Vi=NRTi/Pi). I feel this method is wrong though. Any other suggestions to solving this?

[Enthalpy]

With the information provided here, you must indeed make some assumption about the gas to compute the adiabatic expansion, that is, about its compressibility factor, or equivalently, its heat capacity.

If its U or Cv is 1.5*R then the volume ratio is the temperature ratio (=3) power 1.5. If it's 2.5 (air, nitrogen, diatomic molecules) then the volume ratio is the temperature ratio power 2.5 [Please multiply or divide adequately]. So the final volume needs this information.

This holds for perfect gas, and gas at high pressure like 350 bar for hydraulic bladders are not perfect, for instance.

[orgo814]

Ok I get the method you used. I always thought that the ratio rule only held true for a reversible adiabetic expansion. Is the fact that were returning back to initial conditions signify it's reversible?

[Enthalpy]

Back to the original temperature, not pressure and volume...

Anyway, even a closed cycle, where the fluid regains its initial state, does NOT mean reversible!

But right, you also need "revesible" to compute a volume ratio from the temperature ratio.