A syringe contains 15 mL of air at a pressure of 1.3 atm. What is the new pressure if the plunger is pushed in, so the volume of air is 9 mL?
回答 (3)
2017-07-25 9:37 pm
Consider the air in the syringe.
Initial: P₁ = 1.3 atm, V₁ = 15 mL
New: P₂ = ? atm, V₂ = 9 mL
For a fixed amount of gas at constant temperature: P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
New pressure, P₂ = (1.3 atm) × (15/9) = 2.2 atm (to 2 sig. fig.)
Initial: P₁ = 1.3 atm, V₁ = 15 mL
New: P₂ = ? atm, V₂ = 9 mL
For a fixed amount of gas at constant temperature: P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
New pressure, P₂ = (1.3 atm) × (15/9) = 2.2 atm (to 2 sig. fig.)
2017-07-25 9:39 pm
At constant temperature
P1V1 = P2V2
P1*9mL = 1.3atm * 15mL
P1 = 1.3atm * 15mL / 9mL
P1 = 2.2atm
P1V1 = P2V2
P1*9mL = 1.3atm * 15mL
P1 = 1.3atm * 15mL / 9mL
P1 = 2.2atm
2017-07-25 9:39 pm
Consider that the other factors (temperature, number of gas molecules) are constant, then we can use boyle's law to calculate. P1V1=P2V2.
Thus the answer should be 15*1.3/9=2.167 atm
Thus the answer should be 15*1.3/9=2.167 atm
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