A pump contains 05 L of air at 203 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads 50.8 kPa. What is the new volume of air in the pump? P_(1)V_(1)=P_(2)V_(2) P_(1)=203kPa V_(1)=0.5L P_(2)=50.8kPa V_(2)=? Hint: there is an example problem in your student notes. (Boyle's Law) 4.0 L 0.2 L 1.0 L 2.0 L

Question

A pump contains 05 L of air at 203 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads 50.8 kPa. What is the new volume of air in the pump? P_(1)V_(1)=P_(2)V_(2) P_(1)=203kPa V_(1)=0.5L P_(2)=50.8kPa V_(2)=? Hint: there is an example problem in your student notes. (Boyle's Law) 4.0 L 0.2 L 1.0 L 2.0 L

Answer 1

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional, as long as the temperature remains constant. The formula for Boyle's Law is:P1V1 = P2V2Given information:- Initial pressure (P1) = 203 kPa- Initial volume (V1) = 0.5 L- Final pressure (P2) = 50.8 kPa- Final volume (V2) =?Substituting the given values into the Boyle's Law formula:P1V1 = P2V2203 kPa × 0.5 L = 50.8 kPa × V2V2 = (203 kPa × 0.5 L) / 50.8 kPaV2 = 2.0 LTherefore, the new volume of air in the pump is 2.0 L.

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