The apparatus shown consists of three bulbs connected by stopcocks. What is the pressure inside the system when the stopcocks are opened? Assume that the lines connecting the bulbs have zero volume and that the temperature remains constant.

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$0.412 atm$

$1.412 atm$

$2.412 atm$

$3.412 atm$

answer is B.

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Detailed Solution

This is initially a Boyle’s Law problem, because only P and V are changing while n and T remain fixed. The initial volume for each gas is the volume of their individual bulbs. The final volume for each gas is the total volume of the three bulbs.
$n R T = P_{1} V_{1}; V_{r} = 1.50 + 1.00 + 2.00 = 4.50 L$
$For C O_{2}; P_{f} = \frac{P_{1} V_{1}}{V_{f}} = \frac{(2.13 a t m) (1.50 L)}{(4.50 L)} = 0.710 a t m$
$For H_{2}; P_{f} = \frac{P_{1} V_{1}}{V_{f}} = \frac{(0.861 a t m) (1.50 L)}{(4.50 L)} = 0.191 a t m$
$For A r_{2}; P_{f} = \frac{P_{1} V_{1}}{V_{f}} = \frac{(1.15 a t m) (2.00 L)}{(4.50 L)} = 0.511 a t m$
From Dalton’s Law, $P_{t o t a l} = P_{C O 2} + P_{H 2} + P_{A r}$