answer a.b and c a Secure hitpal/session.manteringchemistry.com/myctfitemView as

Question

answer a.b and c

a Secure hitpal/session.manteringchemistry.com/myctfitemView assignmemPreblemID-88100672 Probiem 8.35 A sample af helure gas has a vekume of 6 50 L at a presure of 845 mmlg and a tempessture of 25 Whot is the pressare al the gas in atm when the wahme and tenperafure of the pas sampla ane changed to 1560 ml and 340 Kif the amount of gas is corstant? Incorrect: Try Again; 4 attempts remaining What is the pressase of the gas in atm whes the voka and changed to 227 Lase 15 CEthe amount af gas e consant? Incerrect; Try Again; 4 attempts cemaining Pan c h th·presuns! theOasi" atm wh" the vtme·rd tempatara" of lbe gn changed to 1381rd 4,rthe anan'dan a cassaer

Explanation / Answer

part A

from ideal gas equation

P1V1/T1 = P2V2/T2

P1 = 845 mmHg = 845/760 = 1.11 atm

V1 = 6.5 L

T1 = 25 c = 298 k

P2 = ?

v2 = 1.56 L

T2 = 340 K

(1.11*6.5/298) = (P2*1.56/340)

P2 = 5.28 atm

part B

from ideal gas equation

P1V1/T1 = P2V2/T2

P1 = 845 mmHg = 845/760 = 1.11 atm

V1 = 6.5 L

T1 = 25 c = 298 k

P2 = ?

v2 = 2.27 L

T2 = 288.15 K

(1.11*6.5/298) = (P2*2.27/288.15)

P2 = 3.07 atm

part C

from ideal gas equation

P1V1/T1 = P2V2/T2

P1 = 845 mmHg = 845/760 = 1.11 atm

V1 = 6.5 L

T1 = 25 c = 298 k

P2 = ?

v2 = 13.8 L

T2 = 45+273.15 = 318.15 K

(1.11*6.5/298) = (P2*13.8/318.15)

P2 = 0.56 atm

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