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A 22.0 cm³ sample of oxygen gas expands to -78.0 cm³. If the original pressure was 570 mmHg, what is the new pressure in atm? (Round to 3 digits.)

User Mneri
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Final answer:

The new pressure of the expanded 22.0 cm³ sample of oxygen gas can be calculated using Boyle's Law and then converted to atm, after correcting the typographical error regarding the final volume.

Step-by-step explanation:

To find the new pressure in atmospheres when a 22.0 cm³ sample of oxygen gas expands to -78.0 cm³, we must first correct the provided expansion value, as negative volumes are not physically meaningful. Assuming this was a typographical error and that the volume expanded to 78.0 cm³, we can use Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume (P1V1 = P2V2).

Given:
Original Volume (V1) = 22.0 cm³
Final Volume (V2) = 78.0 cm³
Original Pressure (P1) = 570 mmHg
The final pressure (P2) can then be calculated as follows:

P1V1 = P2V2
570 mmHg × 22.0 cm³ = P2 × 78.0 cm³
So, P2 = (570 mmHg × 22.0 cm³) / 78.0 cm³

After calculating P2 in mmHg, we convert it to atm using the conversion factor (1 atm = 760 mmHg).

P2 (in atm) = P2 (in mmHg) / 760

After performing these calculations, we express the final pressure P2 in atmospheres, rounded to three decimal places as requested. It is important to note that the initial volume value should be positive as gas volumes cannot be negative.

User TranquilMarmot
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