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Sam's bike tire contains 15 units of air particles and has a volume of 160mL. Under these conditions the pressure reads 13 psi. The tire develops a leak. Now it contains 10…

Sam's bike tire contains 15 units of air particles and has a volume of 160mL. Under these conditions the pressure reads 13 psi. The tire develops a leak. Now it contains 10…
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Sam's bike tire contains 15 units of air particles and has a volume of 160mL. Under these conditions the pressure reads 13 psi. The tire develops a leak. Now it contains 10 units of air particles and has contracted to a volume of 150mL. What would the tire pressure be now?

User Fresco
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1 Answer

6 votes
Use ideal gas equation, with T constant.

pV =nRT => pV / n = RT = constant

n = K* [units of particles]

pV / [units of particles] = constant

13 psi * 160 mL / 15 units = p * 150 mL / 10 units =>

=> p = [13psi*160mL/15units]*[10units/150mL] = 9.2 psi


User Chris Page
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