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When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is 4.00 atm, is opened, what will be the final pressure in the two bulbs

User Tmoasz
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1 Answer

6 votes

Answer:


P_(C) = 3.2\, atm

Step-by-step explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:


P_(A) \cdot V_(A) = n_(A) \cdot R_(u) \cdot T

Bulb B (3 L, 4 atm) - Before opening:


P_(B) \cdot V_(B) = n_(B) \cdot R_(u) \cdot T

Bulbs A & B (5 L) - After opening:


P_(C) \cdot (V_(A) + V_(B)) = (n_(A) + n_(B))\cdot R_(u) \cdot T

After some algebraic manipulation, a formula for final pressure is derived:


P_(C) = (P_(A)\cdot V_(A) + P_(B)\cdot V_(B))/(V_(A)+V_(B))

And final pressure is obtained:


P_(C) = ((2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L))/(5\,L)


P_(C) = 3.2\, atm

User NGR
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