Question #f7b80

1 Answer
Ernest Z.
Jun 7, 2014

This is an example of a Boyle's Law problem.

#P_1V_1=P_2V_2#

#P_1# = 1.00 atm; #V_1# = 0.500 L
#P_2# = 3.25 atm; #V_2# = ?

#V_2 = V_1 × P_1/P_2 # = 0.500 L × #(1.00"atm")/(3.25"atm")# = 0.154 L

This makes sense. The pressure has increased about 3-fold, so the volume decreases to ⅓ of its original value.

#0.50/3# = 0.17

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