If the temperature is held constant during this process and the final pressure is 683 torrtorr, what is the volume of the bulb that was originally filled with gas?

Let the volume of the unknown bulb = X L

The volume of the system, after opening valve = (X + 0.72 L)

Use Boyles law gas equation,

P1V1 = P2V2 (at temperature is constant)

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg (torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987 * (X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L