Answer:
[tex]P_{C} = 3.2\, atm[/tex]
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:
[tex]P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T[/tex]
Bulb B (3 L, 4 atm) - Before opening:
[tex]P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T[/tex]
Bulbs A & B (5 L) - After opening:
[tex]P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T[/tex]
After some algebraic manipulation, a formula for final pressure is derived:
[tex]P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}[/tex]
And final pressure is obtained:
[tex]P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}[/tex]
[tex]P_{C} = 3.2\, atm[/tex]
