Ideal Gas Law Problem: Balloon Volume Change @ Different Pressures

[PennyGirl]

Homework Statement

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?

Homework Equations

P*V=n*R*T

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?

 

[GunnaSix]

Then, I said that...
P1*V1=P2*V2

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

 

[PennyGirl]

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

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?

 

[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.

 

[Borek]

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.

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

 

[PennyGirl]

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.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

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...

 

[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?

 

[vela]

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?

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

 

[PennyGirl]

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

 

[vela]

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.

 

[PennyGirl]

Thanks a ton! I had forgotted about that equation...