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.

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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