Question

The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant?

Answer 1

Using Boyle's law, `P_1V_1=P_2V_2`, `P_2~=0.40*atm`.

Explanation:

From Boyle's law, `P_1V_1=P_2V_2`.

`P_2 = (P_1V_1)/V_2` `=` `(1.20*atmxx2.25*cancelL)/(6.50*cancelL)` `=` `??atm`