# Figure 511 5 21 sample problem 52 applying the volume

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Figure 5.11
Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyles apprentice finds that the air trapped in a J tube occupies 24.8 cm3at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? 2
5-22 Sample Problem 5.2 SOLUTION: P1= 1.12 atm V1= 24.8 cm3P2= 2.64 atm V2= unknownnand Tare constant = 0.0248 L P1V1n1T1P2V2n2T2=P1V1= P2V2= 0.0248 L x 1.12 atm 2.64 atm = 0.0105 L 24.8 cm3x 1 mL 1 cm3L 103 mL x V2= P1P2V1x
Sample Problem 5.3 Applying the Volume-Temperature and Pressure-Temperature Relationships PROBLEM: A balloon is filled with 1.95 L of air at 25ºC and then placed in a car in the sun. What is the volume of the balloon when the temperature in the car reaches 90ºC?
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5-24 SOLUTION: Summarizing the gas variables: Converting T from ºC to K: T1(K) = 25ºC + 273.15 = 298 KT2(K) = 90ºC + 273.15 = 363 KV1= 1.95 LV2= 1.95 LT1= 25ºC (convert to K)T2= 90ºC (convert to K) Pand nremain constant Sample Problem 5.3 ºC + 273.15 = KT1and T2(ºC) T1and T2(K) multiply by T2/T1V1(L) V2(L)
5-25 Sample Problem 5.3 SOLUTION (continued): Rearranging the generalized ideal gas equation and solving for V2: At fixed nand P, we have P1V1n1T1P2V2n2T2= V1T1V2T2= or V2= V1x T2 T1 = 1.95 L x 363 K 298 K = 2.38 L CHECK:Let’s predict the change to check the math: Because T2> T1, we expect V2> V1. Thus, the temperature ratio should be > 1 (T2in the Numerator). The Tratio (363/298) is about 1.2, so the Vratio should also be about 1.2 (2.4/2.0).
5-26 Sample Problem 5.4 Applying the Volume-Amount RelationshipPROBLEM:A scale model of a blimp rises when it is filled with helium to a volume of 55.0 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant Tand P.
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5-27 Sample Problem 5.4 SOLUTION: n1= 1.10 mol V1= 26.2 dm3n2= unknown V2= 55.0 dm3Tand Pare constant P1V1n1T1P2V2n2T2== 1.10 mol x 55.0 dm326.2 dm3= 2.31 mol He n2= V2V1n1xV1n1V2n2=Additional amount of He needed = 2.31 mol – 1.10 mol = 1.21 mol He 1.21 mol He x 4.003 g He 1 mol He = 4.84 g He