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An oxygen container has a volume of 48 mL and a pressure of 420 kPa. What is the volume of this gas when the pressure is 105 kPa?

A) 192 mL
B) 200 mL
C) 180 mL
D) 168 mL

User KZoNE
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1 Answer

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Final answer:

By applying Boyle's Law, the new volume of oxygen gas when the pressure changes from 420 kPa to 105 kPa is calculated to be 192 mL, as pressure and volume are inversely proportional in a closed system at constant temperature.

Step-by-step explanation:

To solve for the new volume of oxygen gas when the pressure changes, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature and the amount of gas are held constant. The formula for Boyle's Law is P1V1 = P2V2, where P represents pressure and V represents volume, while the subscripts 1 and 2 refer to the initial and final states, respectively.

In the given problem, the initial pressure P1 is 420 kPa, the initial volume V1 is 48 mL, and the final pressure P2 is 105 kPa. We are asked to find the final volume V2. By rearranging the formula to solve for V2, we get V2 = (P1V1) / P2.

Inserting the known values, we have V2 = (420 kPa × 48 mL) / 105 kPa = 20160 kPa·mL / 105 kPa = 192 mL. Therefore, the new volume of the gas when the pressure is 105 kPa is 192 mL.

User Gerd Klima
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