A balloon contains 3.90 m3 of argon gas. It is slowly heated at a constant temperature so that its volume triples. If the gas does 3.00 kJ of work in the expansion, what was the initial pressure in the balloon? Isothermal Change of an Ideal Gas: W= -nRT(ln Vf/Vi) PiVi=PfVf= nRT Adiabatic Change of an Ideal Gas: Pi(Vi^1.67)= Pf(Vf^1.67) Dealing with Argon which is a monoatomic gas hence the 1.67 PiVi/Ti= PfVf/Tf I plugged my values into the isothermal equation: 3.00kJ= -nRT(ln 11.7/3.90) and solved for nRT. The 11.7 is my final volume because it triples, right? I got 2.73071768. Then, based on the PiVi= nRT equation I solved for Pi. Pi(3.90)= 2.73071768 and got Pi= 0.70018402. This was incorrect. Then, I tried solving for Pi using the Adiabatic Change equation: Pi(3.90^1.67)= 2.73071768 and got Pi= 0.281318899. This was also incorrect. Where am I going wrong?