Finding the Initial Pressure of A Balloon

[chunkytuna21]

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?

 

[rl.bhat]

Convert kJ to J.

 

[chunkytuna21]

I tried that already. I'm thinking it's either my calculations or my approach to the problem. I find it kind of weird that it could be both an isothermal and adiabatic change, but what do I know?

 

[rl.bhat]

In the problem you have to find out number of moles of argon gas.
Density of argon = 1.784 kg/m^3
Molecular weight = 40.

 

[chunkytuna21]

Ok, I found the moles of argon, but that still doesn't give me the right answer.

3.90m^3 of argon * 1.784= 6.9576kg
6.9576kg * 1000= 6957.6g
6957.6g/40g/mol= 173.94mol

Plugged it into the equation:
3kJ or 3000J (doesn't work either way)= (173.94)(8.314)(T)(ln3)
Solved for nRT by 3kJ/ln3= 2.73071768
PiVi= nRT
Pi(3.90)= 2.73071768
Pi= 0.700MPa or 700184Pa, both are wrong.

I even tried it the other way:
Pi(3.90^1.67)= 2.73071768
Pi= 0.281MPa or 281318Pa, also wrong.

This problem doesn't make any sense and my answers came out the same as they did in my first post even with the moles.

 

[rl.bhat]

Pi*Vi = nRT. So
Pi = nRT/V
= 173.94*2731/3.9
= ?

 

[chunkytuna21]

= 121802.6, but that's still wrong. 2731 should be 2.731. It doesn't really matter though because I did it both ways.

 

[rl.bhat]

What is the answer?

 

[chunkytuna21]

I don't know the answer; my online homework gives me a set amount of trials. I don't get the answer until I have used up all the trials or solve it myself. I've still got 10 trials left so I won't know the answer for awhile, but according to your suggestion the initial pressure should be 121803. I still need to know how to solve the problem because it might show up on my exam.

 

[chunkytuna21]

Any ideas?