**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 H__

__2__

__0__

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

ReplyDeleteWell 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!

ReplyDelete