What is the final pressure in the balloons after being disconnected?

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After disconnecting the balloons, the final pressure can be calculated using the formula p = (2p1p2) / (p1 + p2), resulting in a pressure of 166.67 kPa. The volumes of the balloons do not factor into this equation because the mass of gas and its properties remain constant. It is important to treat the masses of the gas in each balloon as different, while the molar mass can be considered the same. Some participants suggested exploring the mass flow rate, but the focus should remain on the final pressure after the balloons are disconnected. The discussion emphasizes understanding the ideal gas law and the conditions for equilibrium after disconnection.
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Homework Statement


Two balloons are connected by a faucet.
Gas in the first balloon is at pressure p_1=100kPa, and in the second is p_2=0,5MPa.
Volumes are V_1=0,12m^3 and V_2=0,5m^3.
Temperature of a gas is constant.
Find pressure in balloons p after faucet is opened (balloons are not connected).

Homework Equations


pV=\frac{m}{M}RT - state of ideal gas

The Attempt at a Solution


State of gas before before balloons are disconnected is
p_1V_1=\frac{m}{M}RT,p_2V_2=\frac{m}{M}RT
and after balloons are disconnected is
p(V_1+V_2)=\frac{2mRT}{M}\Rightarrow p=\frac{2p_1p_2}{p_1+p_2}=166,67kPa.

Is this correct? How volumes V_1 and V_2 are not relevant in the equation for p?
 
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gruba said:
##p_1V_1=\frac{m}{M}RT,p_2V_2=\frac{m}{M}RT##
Are m and M the same for both?
 
haruspex said:
Are m and M the same for both?
Mass of a balloon m and atomic mass M are not given. Could you elaborate how to set the equations
for finding the pressure p in second case (disconnected balloons)?
 
gruba said:
Mass of a balloon m and atomic mass M are not given. Could you elaborate how to set the equations
for finding the pressure p in second case (disconnected balloons)?
You can assume M is the same for both, but not m. Assign two different unknowns. You have enough equations to cope with that.
 
I'm not sure as I'm only just learning this stuff myself though it appears I'm a couple of weeks behind you but wouldn't it be worth finding the mass flow rate? Sorry if that's wrong.
 
JustDerek said:
I'm not sure as I'm only just learning this stuff myself though it appears I'm a couple of weeks behind you but wouldn't it be worth finding the mass flow rate? Sorry if that's wrong.
I read the question as asking about the final state, after flow has ceased and temperatures have returned to ambient.
(Otherwise there is not enough information.)
 
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