# Homework Help: Ideal Gas Law Problem

1. Jan 8, 2010

### PennyGirl

1. The problem statement, all variables and given/known data
You have a balloon (stretchy) filled with one liter of He at 1 atm and 298 Kalvin. The balloon is suddenly placed into a room at .5 atm and 298 Kalvin. What is the temperature of the gas inside the balloon a little while after this happens?

2. Relevant equations
P*V=n*R*T

3. The attempt at a solution
So first, I calculated the number of moles of He in the balloon, using the ideal gas law as follows...
(1 atm)*(1 L)=n*.0821*(298 K)
n=.0409 mol

Then, I said that...
P1*V1=P2*V2
(1atm)*(1L)=(.5atm)*(V2)
therefore, V2 = 2L

From there, (using ideal gas law again...)
(.5atm)*(2L)=.0409 mol(.0821)*(T)
T=298K
but this answer didn't make sense to me...shouldn't temperature in the balloon change?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 8, 2010

### GunnaSix

This is only true if the temperature doesn't change! Your work is correct, though; the temperature should stay the same.

3. Jan 8, 2010

### PennyGirl

So...I was thinking at first that I could use P1*V1/T1=P2*V2/T2 so I would get
(1atm)*(1L)/(298K)=(.5atm)*(V2)/T2...but then I have two unknowns?

Then I tried that eqn with
(.5atm)*V2=.0409mol*.0821*T2

But these two equations are not indpendent? Help?

4. Jan 9, 2010

### GunnaSix

You're confusing yourself. The "temperature of the gas inside the balloon a little while after this happens" should be the same as the temperature of the room, or 298K.

5. Jan 9, 2010

### Staff: Mentor

For me wording is ambiguous. "A little while after" doesn't mean "after thermal equilibrium has been achieved". It can mean whatever you want. Say, 5 minutes. How is it related to thermal equilibrium? No idea.

--
methods

6. Jan 9, 2010

### PennyGirl

okay...so if I try and calculate the temp in the balloon "shortly after the action is performed" (exact wording from the problem), it shouldn't be 298K ? I also find this problem ambiguous, because I have no idea whether or not thermal equilibrium is reached, which would happen eventually...

7. Jan 9, 2010

### PennyGirl

Okay...i just realized that in the problem it states that the balloon is insulated...which means it shouldn't be 298 K (right?)...so what if i try an energy balance
I know that change in Q=0, change in W=0, and assume changes in kinetic and potential energy both equal zero, which means the change in entropy = 0
so, H = U + deltaP*V

H= (.5atm)*(1L)

but i don't remember how to calculate delta H, isn't it in a table or something?

8. Jan 9, 2010

### vela

Staff Emeritus
Why do you think the gas in the balloon did no work when it expanded against the 0.5 atm outside?

9. Jan 10, 2010

### PennyGirl

any other ideas on how to approach this problem? I feel like I'm heading in the wrong direction with this

10. Jan 10, 2010

### vela

Staff Emeritus
You said the balloon is insulated, so the gas expands adiabatically. For adiabatic expansions, you have $PV^\gamma=\texttt{constant}$ where $\gamma=c_p/c_v$.

11. Jan 10, 2010