Chem help please!!

Question

1) A sample of helium gas has a volume of
302 mL at 0.52 atm. What pressure is needed
to reduce the volume at constant temperature
to 69 mL?
Answer in units of atm

 

2)A gas has a pressure of 2.35 atm and occupies
a volume of 5.7 L. If the gas is compressed to
a volume of 1.3 L, what will its pressure be,
assuming constant temperature?
Answer in units of atm.

 

3)The volume of a gas at 24◦C and 0.39 atm
is 91 mL. What volume will the same gas
sample occupy at standard conditions?
Answer in units of mL.

Answer 1

1)  V1 = 302 ml 

P1 = 0.52 atm

P2 = ?

V2 = 69 ml 

 

Boyle's Law

P1V1 = P2V2

P2 = P1V1/V2

P2 = (0.52 atm)(302 ml)/69 ml

P2 = 2.3 atm (to 2 significant figures)

 

2)  P1 = 2.35 atm

V1 = 5.7 L

P2 = ?

V2 = 1.3 L

Boyle's Law

P1V1 = P2V2

P2 = (2.35 atm)(5.7 L)/1.3 L

P2 = 10.3 atm = 10. atm (to 2 significant figures)

 

3)There are two ways to do this problem:

 

V1 = 91 ml

P1 = 0.39 atm

T1 = 24ºC + 273 = 297ºK

V2 = ?

P2 = 1 atm

T2 = 273ºK

P1V1/T1 = P2V2/T2

V2 = P1V1T2/T1P2 = (0.39)(91)(273)/(297)(1)

V2 = 32.6 ml = 33 mls (to 2 significant figures)

 

The other way (easier) is:

PV = nRT and solve for n:

n = PV/RT = (0.39 atm)(0.091 L)/(0.08206 Latm/Kmol)(297K)  .... NOTE:  volume is converted to liters

n = 0.00146 moles

At STP 1 mole = 22.4 L

0.00146 moles x 22.4 L/mole = 0.0326 liter = 32.6 mls = 33 mls (to 2 significant figures)