Homework Answers
Ans 19 : a) 11.3 mL
Using the relation P1V1T2 = P2V2T1
P1 = 659 torr = 0.867105 atm
V1 = 14.1 mL = 0.0141 L
T1 = 22oC = 295.15 K
at STP :
P2 = 1.00 atm
T2 = 273.15 atm
putting values in the relation :
0.867105 atm x 0.0141 L x 273.15 K = 1.00 atm x V2 x 295.15 K
V2 = 0.0113 L
volume in mL = 0.0113 x 1000 mL = 11.3 mL
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question in first picture
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