Tuesday, May 17, 2011

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: V/T1 = V2/T2

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

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)
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)
g of O2= 20.0g
V of H2O=?
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


  1. for part c when you convert the temperature into kelvin it think its 498, not 489

  2. Well done Sam! The way you bolded answers and used colors to help organize your blog was VERY helpful and easy to follow. I got all the correct answers that you got except part c in which i got 489 K. I might go back and check to see if that was right. Besides that... great job! Maybe even next time you could just math type to make things a little neater for the longer equations and such. Overall awesome!