Question

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

Answer 1

Answer:

$P_{C} = 3.2\, atm$

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:

$P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}$

And final pressure is obtained:

$P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}$

$P_{C} = 3.2\, atm$