Warmup:1. The volume of a gas is 27.5 mL at 22.0°C

and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm?

2. Suppose that 120. mL of argon is collected over water at 25°C with an atmospheric pressure of 780 mmHg. Compute the volume that this argon gas would occupy at STP instead.

1. The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm?

Initial After

P1 = P2 =

V1 = V2 =

T1 = T2 =

P1T1

P2T2

0.974 atm 0.993 atm

27.5 mL ?

22°C 295 K

15°C 288 K

2

22

1

11

TVP

TVP

V2 =

atm0.993 X K 295

K 288 X mL 27.5 X atm0.974

V2 = 26.3 mL

V1

V2

21

211

PTTVP

2. Suppose that 120. mL of argon is collected over water at 25°C with an atmospheric pressure of 780 mmHg Compute the volume that this argon gas would occupy at STP instead.

Part 1: What is the pressure of argon INITIALLY?

Part 2: What is the volume AFTER I change the conditions?

Ptotal = PAr + PH2O

780mmHg = PAr + 23.8mmHg

PAr = 756 mmHg

Initial FinalP1 = P2 =

V1 = V2 =

T1 = T2 =

756 mmHg

120 mL

298 K

760 mmHg

?273K

2

22

1

11

TVP

TVP

21

211 2

PTTVP

V

Answer: 109 mL

The Ideal Gas Law

The Ideal Gas Law relates pressure, volume, temperature, and moles of a gas :

PV = nRTPressure of Gas

(in atm)Volume of Gas

(in L) Moles of Gas(in moles)

Universal Gas Constant

Kmolatm L 0.0821

Temperature of Gas(in Kelvin)

1. What is the pressure in atmospheres exerted by a 5.00 mol sample of N2 in a 10.0 L container at 298K?

P

V

n

R

T

?10.0 L

5.00 mol

298 KKmolatm L 0.0821

PV = nRT

VnRT

P

L 10.0

K) )(298K molatm L 0.0821

mol)( (5.00 P

P = 12.2 atm

2. What pressure (in kPa) will be exerted by 0.125 moles of neon gas contained in a 9.22 L vessel at 25.0°C?

P

V

n

R

T

9.22 L?

0.125 mol

298 KKmolatm L 0.0821

PV = nRT

VnRT

P

L 9.22

K) )(298K molatm L 0.0821

mol)( (0.125 P

P = 0.332 atmHint: 1 atm = 101.3 kPa

P = 33.6 kPa

3. How many grams are contained in 5.00 L of chlorine gas at STP?

P

V

n

R

T

5.00 L1.00 atm

?

273 KKmolatm L 0.0821

PV = nRT

RTPV

n

)(273K)K molatm L 0.0821

(

L) atm)(5.00 (1.00 n

0.223 mol Cl2

mol Cl2

g Cl2

170.9

15.8 g Cl2

4. C2H4(l) + 2O2(g) 2CO2(g) + 2H2O(g) A 6.82 x 10-3 mole sample of ethene goes through complete combustion,

producing carbon dioxide gas in a 250. mL container at 400 K. What is the pressure exerted by the gas?

P

V

n

R

T

0.250 L?

400 KKmolatm L 0.0821

VnRT

P 250 mL ? L

6.82x10-3 mole C2H4

mole C2H4

mol CO2

12

2 atm

0.0136 mol

P = (1.36 x 10-2)(0.0821)(400)0.250

= 1.36 x 10-2 moles CO2

5. A 32.5 gram sample of an unknown diatomic gas was placed in a 250. mL container at 400. K and 60.2 atm. What is the molar mass, and identity of this gas?

P

V

n

R

T

0.250 L60.2 atm

400 KKmolatm L 0.0821

PV = nRT

71 g/mole

?(60.2)(0.250) = n(o.0821)

(400.)n = 0.458 moles

molar mass = g/mole

molar mass = 32.5 g/0.458mole

71 / 2 diatomic = 35.5 = Cl2

R lab!Stuff to think about:1. This is a gas lab. Vinegar and baking soda aren’t gases. What? 2. What are we trying to DO? I know this is an experiment with lots of steps and stuff, but what are we trying to PROVE?3. How will you collect data for each variable (P,V,n,T)? Specifically?4. Why did I give you a balloon? 5. Which chemical will be the limiting reactant in your reaction?6. How important is it that all of your baking soda is used up in the reaction? What about the vinegar?7.How do you KNOW that your yield is enough?8. How do you KNOW that your R value is accurate?9. Barometric pressure on the website is in inches of Hg, you must convert!