Secure htt/session masteringchemistry com/myct/temView assignmentProblemiD-87713

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Secure htt/session masteringchemistry com/myct/temView assignmentProblemiD-877133868viewprintboffset-next NCHS Changes in Temperature ± Changes in Temperature The iSeal gas law (PV- n.RT) desoribes the relationship among pressure P. volume V, temperature T, and molar amount n When n and V are fixed, the equation can be rearranged to take the following form where k is a constant When n and P are foxed, the expression becomes Part A The pressure inside a hydrogen-iled container was 2 10 atm at 21 C. What would the pressure be t the container was heated to 90 C Express your answer with the appropriate units You did not open hints for this part ANSWER Part B Atstandard le peatre "d pressure (0-cand 1 00atm 1.00mol of an ideal gas o Express your answer with the appropriate units You did not open hints for this part ANSWER apom a volume of 22.4 what voling would the same amortofgas ocopy at the san e pressue and 85 Type here to search

Explanation / Answer

Q1) Part A

(P/T) initial = (P/T) final

2.10atm/ 294K = P2/363K

Thus final pressure at 363K (90C) = 2.592 atm

Part B

(V/T) initial = (V/T) final

22.4L / 273K = V2 / 358K

Thus final volume at 358K (85C) is 29.374L

Q2)

Part A)

Only second and fourth options are correct for Boyles law.

For a fixed mass of gas, pressure is inversely proportional to Volume at constant temperature (option2)

For a fixed mass of gas, volume is inversely proportional to pressure at constant temperature (option4)

Part B)

ballon filled at STP and released to air at0.5 atm, Volume increases.

ballon filled at STP and submerged into water at 1.25 atm pressure , VOLUME DECREASES.

ballon filled under water at 1.15atm but released into air at STp, VOLUME INCREASES

ballon filled at STP and released into air at 1 atm, NO CHANGE IN VOLUME

Part C

k1 = PVinitial

=13.0L x 3.0 atm

= 39 L. atm

k2 = PV final

= 6.0atmx 6.5 L

= 39 L.atm

Thus k1 = k2

and the gas follows Boyle's law

PartD

From Boyles law

PV initial = PV final

15L x 7.5atm = P x 3.8L

Thus final pressure =29.60 atm

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