Exam Review: Question #24

24. Gas Laws:

a. A gas has a volume of 590.0 mL at a temperature of -55.0 degrees C. What volume will the gas occupy at 30.0 degrees.

1. This problem requires Charles' Law: V1 /T1 = V2/T2

Data:

V1= 0.590 L

T1= -55.0 C + 273.15 K = 218.2 K

V2= ?

T2= 30.0 C +273.15 K =303.2 K

Formula:

V2=V1(T2)/T1 V2=(0.590L)(303.2K)/(218.2K)

V2=0.819 L

b. A gas occupies a volume of 140.0 mL at 35.0 degrees Celcius and 97 kPa. What is the volume of the gas at STP (standard temperature and pressure)?

*requires the combined gas law

Data: Formula:

V1= 0.140L P1V1/T1 = P2V2/T2

T1=35.0 C + 273.15 K= 308.2K  (97 kPa)(0.140 L)/308.2 = V2(101.325 kPa)/273.15 K

P1= 97kPa  V2= (97)(0.140)(273.15)/(101.325)(308.2)

V2=?

T2=273.15K V2= 0.12 L

P2=101.325 kPa

c. If 20.0g of O2 is used in a combustion reaction with excess hydrogen, what volume of water vapor will be produced at a pressure of 825mmHg and 225 degrees Celcius?

2H2 (g) + O2 (g) > 2H2O (g)

Data:

g of O2= 20.0g

V of H2O=?

P=825mmHg

T=225 C +273.15 K= 498K

Step 1: convert 825 mmHg to atm

825mmHg x 1 atm/760.0 mmHg = 1.09 atm

Step 2: convert grams of oxygen gas > moles of oxygen gas > moles of H2O

20.0g O2 x 1 mol/32g x 2 mol H2O/ 1 mol O2 = 1.3 mol H2O

Step 3: Solve for Volume of water vapor

(use ideal gas law> PV=nRT)

V=(1.3 mol H2O)(.0821Latm/molK)(489 K)/1.09atm

V=48.8 L H20