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Chapter 11 Homework 11.46 A 0.48 mole sample of helium gas

Chapter 11 Homework
11.46 A 0.48 mole sample of helium gas occupies a volume of 11.7 L. What is the volume of 0.72 mol of
helium gas under the same conditions?
Solution:
It’s a gas! PV=nRT
Easiest thing is to collect things that are constant on one side and only look at the things that are
changing.
𝑃𝑉 = 𝑛𝑅𝑇
The P and T must be constant. So only V and n are changing, so…
𝑃𝑉
= 𝑅𝑇
𝑛
𝑉 𝑅𝑇
=
= π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑛
𝑃
So, as long as P and T stay constant, V/n will stay constant:
π‘‰π‘π‘’π‘“π‘œπ‘Ÿπ‘’ π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ
=
π‘›π‘π‘’π‘“π‘œπ‘Ÿπ‘’ π‘›π‘Žπ‘“π‘‘π‘’π‘Ÿ
π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ
11.7 𝐿
=
0.48 π‘šπ‘œπ‘™ 0.72 π‘šπ‘œπ‘™
π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ = 17.55 𝐿
11.54 A bag of potato chips contains 585 mL of air at 25 C and a pressure of 765 mm Hg. Assuming the
bag does not break, what will be its volume at the top of a mountain where the pressure is 442 mm Hg
and the temperature is 5 C.
Solution:
There’s GASES! It’s an Ideal Gas Law problem. There’s 2 ways to do this:
One is to focus on just what is changing: P, V, and T are changing so n is constant
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉
= 𝑛𝑅 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑇
Since PV/T is constant:
π‘ƒπ‘π‘’π‘“π‘œπ‘Ÿπ‘’ π‘‰π‘π‘’π‘“π‘œπ‘Ÿπ‘’ π‘ƒπ‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ
=
π‘‡π‘π‘’π‘“π‘œπ‘Ÿπ‘’
π‘‡π‘Žπ‘“π‘‘π‘’π‘Ÿ
T has to ALWAYS be in Kelvins:
𝐾 = ℃ + 273.15
𝐾 = 5℃ + 273.15 = 278.15 𝐾
𝐾 = 25℃ + 273.15 = 298.15 𝐾
P and V units won’t matter here as long as they are the same β€œbefore” and β€œafter”
(765 π‘šπ‘š 𝐻𝑔)(585 π‘šπΏ) (442 π‘šπ‘š 𝐻𝑔)π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ
=
298.15 𝐾
(278.15 𝐾)
1501
π‘šπ‘šπ»π‘” × π‘šπΏ
π‘šπ‘šπ»π‘”
= 1.589
(π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ )
𝐾
𝐾
π‘‰π‘Žπ‘“π‘‘π‘’π‘Ÿ =
1501
π‘šπΏ = 944.6 π‘šπΏ
1.589
The other way to do it is to use PV=nRT to find moles (β€œn”) initially and then use the whole PV=nRT to
find the new volume. In this case, all my units need to match the units of R so they cancel.
𝑅 = 0.082058
765 π‘šπ‘š 𝐻𝑔
𝐿 π‘Žπ‘‘π‘š
π‘šπ‘œπ‘™ 𝐾
1 π‘Žπ‘‘π‘š
= 1.007 π‘Žπ‘‘π‘š
760 π‘šπ‘š 𝐻𝑔
585 π‘šπΏ
1𝐿
= 0.585 𝐿
1000 π‘šπΏ
𝑃𝑉 = 𝑛𝑅𝑇
(1.007 π‘Žπ‘‘π‘š)(0.585 𝐿) = 𝑛 (0.082058
𝐿 π‘Žπ‘‘π‘š
)(298.15 𝐾)
π‘šπ‘œπ‘™ 𝐾
𝑛 = 0.0241 π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘”π‘Žπ‘ 
The bag stays sealed, so the amount of gas should stay the same.
𝑅 = 0.082058
442 π‘šπ‘š 𝐻𝑔
𝐿 π‘Žπ‘‘π‘š
π‘šπ‘œπ‘™ 𝐾
1 π‘Žπ‘‘π‘š
= 0.582 π‘Žπ‘‘π‘š
760 π‘šπ‘š 𝐻𝑔
𝑃𝑉 = 𝑛𝑅𝑇
(0.582 π‘Žπ‘‘π‘š)(𝑉) = (0.0241 π‘šπ‘œπ‘™π‘’π‘ ) (0.082058
𝐿 π‘Žπ‘‘π‘š
)(278.15 𝐾)
π‘šπ‘œπ‘™ 𝐾
𝑉 = 0.946 𝐿 π‘œπ‘Ÿ 946 π‘šπΏ
11.90 Consider the chemical reaction:
2 H2O (l) β†’ 2 H2 (g) + O2 (g)
How many moles of H2O are required to form 1.3 L of O2 at a temperature of 325 K and a pressure of
0.988 atm?
SOLUTION:
Well, it’s a gas, so:
𝑃𝑉 = 𝑛𝑅𝑇
What do I know about oxygen?
P=0.988 atm
V=1.3 L
T= 325 K
n=?
So, I know everything but β€œn”. I can find moles.
𝑃𝑉 = 𝑛𝑅𝑇
(0.988 π‘Žπ‘‘π‘š)(1.3 𝐿) = 𝑛(0.082058
𝐿 π‘Žπ‘‘π‘š
)(325 𝐾)
π‘šπ‘œπ‘™ 𝐾
𝑛 = 0.0482 π‘šπ‘œπ‘™ 𝑂2
So, I need 0.0482 mol O2. I’m getting it from the reaction, so the balanced equation (stoichiometry) tells
me how much water I need!
0.0482 π‘šπ‘œπ‘™ 𝑂2
2 π‘šπ‘œπ‘™ 𝐻2 𝑂
= 0.0964 π‘šπ‘œπ‘™ 𝐻2 𝑂
1 π‘šπ‘œπ‘™ 𝑂2
11.92 Oxygen gas reacts with powdered aluminum according to the reaction:
4 Al (s) + 3 O2 (g) β†’ 2 Al2O3 (s)
How many liters of O2 gas, measured at 782 mm Hg and 25 C, are required to completely react with 2.4
mol of Al?
SOLUTION:
Well, it’s a gas, so:
𝑃𝑉 = 𝑛𝑅𝑇
What do I know about oxygen?
P=782 mmHg
V=?
T= 25 C
n=?
So, I don’t know V or n, BUT…
I know I need enough oxygen to react with 2.4 mol of Al. How is Al related to oxygen? That’s right:
balanced equation (stoichiometry):
2.4 π‘šπ‘œπ‘™ 𝐴𝑙
P=782 mmHg
V=?
T= 25 C
n=1.8 mol O2
3 π‘šπ‘œπ‘™ 𝑂2
= 1.8 π‘šπ‘œπ‘™ 𝑂2
4 π‘šπ‘œπ‘™ 𝐴𝑙
So, I don’t know V, but that’s the only thing I don’t know. I can find it…I just need the units to cancel
properly:
𝑃 = 782 π‘šπ‘š 𝐻𝑔
1 π‘Žπ‘‘π‘š
= 1.029 π‘Žπ‘‘π‘š
760 π‘šπ‘š 𝐻𝑔
𝑇 = 25℃ + 273.15 = 298.15 𝐾
𝑛 = 1.8 π‘šπ‘œπ‘™
𝑅 = 0.082058
𝐿 π‘Žπ‘‘π‘š
π‘šπ‘œπ‘™ 𝐾
𝑉 =?
𝑃𝑉 = 𝑛𝑅𝑇
(1.029 π‘Žπ‘‘π‘š)(𝑉) = 1.8 π‘šπ‘œπ‘™ (0.082058
𝐿 π‘Žπ‘‘π‘š
)(298.15 𝐾)
π‘šπ‘œπ‘™ 𝐾
𝑉 = 42.8 𝐿
11.94 Sodium reacts with chlorine gas according to the reaction:
2 Na (s) + Cl2 (g) β†’ 2 NaCl (s)
What volume of Cl2 gas, measured at 687 torr and 35 C, is required to form 28 g of NaCl?
Well, it’s a gas, so:
𝑃𝑉 = 𝑛𝑅𝑇
What do I know about chlorine?
P=687 torr
V=?
T= 35 C
n=?
So, I don’t know V or n, BUT…
I do know that I need enough chlorine to make 28 g of NaCl. Back to the reaction!
28 𝑔 π‘π‘ŽπΆπ‘™
1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™ 1 π‘šπ‘œπ‘™ 𝐢𝑙2
= 0 .2395 π‘šπ‘œπ‘™ 𝐢𝑙2
58.453 𝑔 π‘π‘ŽπΆπ‘™ 2 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
P=687 torr
V=?
T= 35 C
n=0.2395 mol Cl2
So, I don’t know V, BUT I can find it. I just need the units to cancel properly:
𝑃 = 687 π‘‘π‘œπ‘Ÿπ‘Ÿ
1 π‘Žπ‘‘π‘š
= 0.904 π‘Žπ‘‘π‘š
760 π‘‘π‘œπ‘Ÿπ‘Ÿ
𝑇 = 35℃ + 273.15 = 308.15 𝐾
𝑛 = 0.2395 π‘šπ‘œπ‘™
𝑅 = 0.082058
𝐿 π‘Žπ‘‘π‘š
π‘šπ‘œπ‘™ 𝐾
𝑉 =?
𝑃𝑉 = 𝑛𝑅𝑇
(0.904 π‘Žπ‘‘π‘š)(𝑉) = 0.2395 π‘šπ‘œπ‘™ (0.082058
𝑉 = 6.70 𝐿
𝐿 π‘Žπ‘‘π‘š
)(308.15 𝐾)
π‘šπ‘œπ‘™ 𝐾

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