Ideal Gas Law II Worksheet by Scorton Creek Publishing - Kevin Cox - Free Printable

Let’s solve each problem step by step using the ideal gas law and related formulas. We’ll use:

Ideal Gas Law:
> PV = nRT

Where:
- P = pressure (in atm, unless otherwise specified)
- V = volume (in liters)
- n = number of moles
- R = gas constant = 0.0821 L·atm/(mol·K) — *unless units differ*
- T = temperature in Kelvin (°C + 273)

Also remember:
- Molar mass of CO₂ = 44 g/mol
- Molar mass of O₂ = 32 g/mol
- Molar mass of N₂ = 28 g/mol
- STP = Standard Temperature and Pressure = 0°C (273 K) and 1 atm → at STP, 1 mole of any gas occupies 22.4 L

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Problem 1: What volume will 2 moles of nitrogen occupy at 720 mm Hg and 20°C?

First, convert units:

- Pressure: 720 mm Hg → convert to atm
→ 760 mm Hg = 1 atm → so 720 / 760 = 0.9474 atm

- Temperature: 20°C → 20 + 273 = 293 K

- n = 2 mol
- R = 0.0821

Use PV = nRT → solve for V:

V = nRT / P
= (2)(0.0821)(293) / 0.9474
= (48.1106) / 0.9474 ≈ 50.78 L

✔ But wait — let’s check if we can use a simpler method with STP conversion? The worksheet shows an answer of 46.2 L — that suggests they may have used different rounding or perhaps assumed something else.

Wait — maybe they used R = 62.4 L·mmHg/(mol·K)? Let’s try that.

If P is in mmHg, then R = 62.4

V = nRT / P = (2)(62.4)(293) / 720
= (36566.4) / 720 ≈ 50.79 L — same as before.

But the worksheet says “(720)(V) = (2)(62.4)(293)” → which gives V = 50.79 L — but then it says “Answer: 46.2 L” — that doesn’t match.

Wait — looking again at the worksheet image description — actually, the user didn’t upload an image; I’m working from text. But in the original prompt, there was a note about “the worksheet shows...” — no, actually, re-reading: the user provided a text representation of a worksheet titled “IDEAL GAS LAW II”, with problems and some answers filled in.

Looking at problem 1 in the text:

> 1. What volume will 2 moles of nitrogen occupy at 720 mm Hg and 20°C?
> (720)(V) = (2)(62.4)(293) → Answer: 46.2 L ← this must be wrong calculation.

Wait — let me recalculate:

(2)(62.4)(293) = 2 × 62.4 = 124.8; 124.8 × 293 = let's compute:

124.8 × 300 = 37,440 minus 124.8 × 7 = 873.6 → 37,440 - 873.6 = 36,566.4

Then 36,566.4 ÷ 720 = ?

720 × 50 = 36,000 → remainder 566.4 → 566.4 / 720 = 0.7867 → total ≈ 50.79 L

So why does the worksheet say 46.2 L? That would be if temperature was 273 K (0°C), not 293 K.

Check: If T = 273 K:

V = (2)(62.4)(273)/720 = (34070.4)/720 ≈ 47.32 L — still not 46.2.

Alternatively, maybe they used R = 0.0821 and forgot to convert pressure?

Try: P = 720 (as if atm) → V = (2)(0.0821)(293)/720 → way too small.

No — perhaps typo in worksheet? Or maybe I misread.

Wait — looking back at the user’s input: the worksheet has handwritten-style answers. For problem 1, it says:

> (720)(V) = (2)(62.4)(293) → Answer: 46.2 L

That math is incorrect. 2×62.4×293 = 36,566.4; divided by 720 is ~50.79, not 46.2.

Unless... did they use T = 273? 2×62.4×273 = 34,070.4 / 720 = 47.32 — still not 46.2.

Or maybe n = 1.8? No.

Perhaps they used R = 62.4 but T = 20+273=293, but miscalculated multiplication.

Let me calculate 62.4 × 293:

60×293 = 17,580
2.4×293 = 703.2
Total = 18,283.2
Times 2 = 36,566.4 — correct.

36,566.4 ÷ 720:

Divide numerator and denominator by 10: 3,656.64 / 72

72 × 50 = 3,600
Subtract: 56.64
72 × 0.7867 ≈ 56.64 → so 50.7867 L

So correct answer should be 50.8 L (rounded).

But since the worksheet says 46.2, and we’re supposed to solve accurately, we go with correct physics.

However — wait! Maybe the problem is using STP correction differently? Let’s see problem 2.

Actually, let’s proceed carefully for all problems, and if the worksheet has errors, we correct them.

But the instruction says: “Solve the problem accurately.” So we do correct calculations.

But let’s double-check problem 1 with another approach.

At STP (0°C, 760 mmHg), 2 moles occupy 2 × 22.4 = 44.8 L.

Now adjust for new conditions:

P1 = 760 mmHg, V1 = 44.8 L, T1 = 273 K
P2 = 720 mmHg, T2 = 293 K, V2 = ?

Combined gas law: P1V1/T1 = P2V2/T2

So V2 = V1 × (P1/P2) × (T2/T1)
= 44.8 × (760/720) × (293/273)

Calculate step by step:

760/720 = 19/18 ≈ 1.0556
293/273 ≈ 1.0733
Product: 1.0556 × 1.0733 ≈ 1.133
Then 44.8 × 1.133 ≈ 50.76 L — same as before.

So definitely 50.8 L is correct.

But the worksheet says 46.2 — perhaps it’s a different problem? Or typo.

Looking ahead, problem 2 also has discrepancy.

To avoid confusion, I’ll solve each problem correctly based on standard chemistry principles.

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Problem 2: What pressure will 25 g of CO₂ exert at 25°C and a volume of 500 mL?

Molar mass CO₂ = 44 g/mol
n = 25 / 44 ≈ 0.5682 mol

T = 25 + 273 = 298 K
V = 500 mL = 0.5 L

R = 0.0821

PV = nRT → P = nRT / V
= (0.5682)(0.0821)(298) / 0.5

First, 0.5682 × 0.0821 ≈ 0.04665
Then × 298 ≈ 13.90
Then / 0.5 = 27.8 atm

Worksheet says: (P)(0.5) = (25/44)(0.0821)(298) → Answer: 27.8 atm — matches!

Good.

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Problem 3: At what temperature will 10 g of O₂ exert at 800 mm Hg and a volume of 400 mL?

Molar mass O₂ = 32 g/mol
n = 10 / 32 = 0.3125 mol

P = 800 mm Hg → use R = 62.4 L·mmHg/(mol·K)
V = 400 mL = 0.4 L

PV = nRT → T = PV / (nR)

T = (800)(0.4) / (0.3125 × 62.4)
= 320 / (19.5) ≈ 16.41 K? That can’t be right — too cold.

Wait — 0.3125 × 62.4 = let's compute:

0.3 × 62.4 = 18.72
0.0125 × 62.4 = 0.78
Total = 19.5 — yes.

320 / 19.5 ≈ 16.41 K — that’s -256°C — impossible for oxygen gas; it would liquefy.

Mistake? Volume is 400 mL = 0.4 L — correct.

Pressure 800 mmHg — high, but possible.

Mass 10g O₂ — 0.3125 mol — reasonable.

But temperature coming out 16K? That suggests error in setup.

Perhaps R value? If we use R = 0.0821, then P must be in atm.

Convert P: 800 mmHg / 760 = 1.0526 atm

Then T = PV / (nR) = (1.0526)(0.4) / (0.3125 × 0.0821)

Numerator: 0.42104
Denominator: 0.3125 × 0.0821 ≈ 0.025656
T = 0.42104 / 0.025656 ≈ 16.41 K — same thing.

This is physically unrealistic. Oxygen boils at 90 K, so at 16K it’s solid.

Probably the volume is 4000 mL? Or mass is 1g? But as given, mathematically it’s 16.4 K.

But worksheet says: (800)(0.4) = (10/32)(62.4)(T) → Answer: 16.4 K — so they got the same.

Perhaps it’s a trick question, or typo in problem. But we’ll report the calculated value.

So T = 16.4 K

But let’s write it as 16 K if rounded.

Worksheet says 16.4 K — ok.

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Problem 4: How many moles of gas occupy 22 L at 24°C and 0.95 atm?

T = 24 + 273 = 297 K
V = 22 L
P = 0.95 atm
R = 0.0821

n = PV / RT = (0.95)(22) / (0.0821)(297)

Numerator: 20.9
Denominator: 0.0821 × 297 ≈ 24.3837
n = 20.9 / 24.3837 ≈ 0.857 mol

Worksheet says: (0.95)(22) = n(0.0821)(297) → Answer: 0.86 mol — matches.

Good.

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Problem 5: What volume will 20 g of argon occupy at 800 torr and 25°C?

Argon molar mass = 40 g/mol
n = 20 / 40 = 0.5 mol

P = 800 torr — note: 1 torr = 1 mmHg, so same as 800 mmHg
Use R = 62.4 L·torr/(mol·K) or convert to atm.

Better to use consistent units.

Convert P to atm: 800 / 760 ≈ 1.0526 atm
T = 25 + 273 = 298 K
R = 0.0821

V = nRT / P = (0.5)(0.0821)(298) / 1.0526

Compute numerator: 0.5 × 0.0821 = 0.04105; × 298 ≈ 12.2329
Divide by 1.0526 ≈ 11.62 L

Using R = 62.4 with P in torr:

V = nRT / P = (0.5)(62.4)(298) / 800

Numerator: 0.5 × 62.4 = 31.2; × 298 = 9297.6
Divide by 800 = 11.622 L

Same.

Worksheet says: (800)(V) = (20/40)(62.4)(298) → Answer: 11.6 L — matches.

Good.

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Problem 6: What pressure will be required to compress 10 g of O₂ at 20°C into a volume of 400 mL?

Molar mass O₂ = 32 g/mol
n = 10 / 32 = 0.3125 mol

T = 20 + 273 = 293 K
V = 400 mL = 0.4 L

Use R = 0.0821 (if P in atm)

P = nRT / V = (0.3125)(0.0821)(293) / 0.4

Compute step by step:

0.3125 × 0.0821 ≈ 0.025656
× 293 ≈ 7.517
/ 0.4 = 18.79 atm

If use R = 62.4 and P in mmHg:

P = nRT / V = (0.3125)(62.4)(293) / 0.4

Numerator: 0.3125 × 62.4 = 19.5; × 293 = 5713.5
/ 0.4 = 14,283.75 mmHg

Convert to atm: 14,283.75 / 760 ≈ 18.79 atm — same.

Worksheet says: (P)(0.4) = (10/32)(0.0821)(293) → Answer: 18.8 atm — matches.

Good.

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Problem 7: At what temperature will a gas of 15 g of He occupy 25 L at 1.5 atm?

Helium molar mass = 4 g/mol
n = 15 / 4 = 3.75 mol

P = 1.5 atm
V = 25 L
R = 0.0821

T = PV / (nR) = (1.5)(25) / (3.75 × 0.0821)

Numerator: 37.5
Denominator: 3.75 × 0.0821 = 0.307875
T = 37.5 / 0.307875 ≈ 121.8 K

Worksheet says: (1.5)(25) = (15/4)(0.0821)(T) → Answer: 122 K — matches.

Good.

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Problem 8: Calculate the pressure in kPa of 755 g of oxygen occupying 1 L at 200°C.

First, molar mass O₂ = 32 g/mol
n = 755 / 32 = 23.59375 mol

T = 200 + 273 = 473 K
V = 1 L

We need P in kPa.

R = 8.314 L·kPa/(mol·K) — because 1 atm = 101.325 kPa, and R = 0.0821 L·atm/(mol·K), so R = 0.0821 × 101.325 ≈ 8.314 L·kPa/(mol·K)

P = nRT / V = (23.59375)(8.314)(473) / 1

First, 23.59375 × 8.314 ≈ let's compute:

20 × 8.314 = 166.28
3.59375 × 8.314 ≈ 3.5 × 8.314 = 29.099; 0.09375 × 8.314 ≈ 0.779; total ≈ 29.878
So total ≈ 166.28 + 29.878 = 196.158

Then × 473 ≈ 196.158 × 473

200 × 473 = 94,600
Minus 3.842 × 473 ≈ 3.8 × 473 = 1797.4; 0.042×473≈19.866; total subtract ≈ 1817.266
So 94,600 - 1,817.266 = 92,782.734 — wait, no:

196.158 × 473 = ?

Better: 196.158 × 400 = 78,463.2
196.158 × 70 = 13,731.06
196.158 × 3 = 588.474
Sum: 78,463.2 + 13,731.06 = 92,194.26; +588.474 = 92,782.734 kPa

That seems very high — 92,783 kPa is about 915 atm — possible for compressed gas, but let's verify.

n = 755/32 = 23.59375 mol — yes, a lot of moles in 1L at high temp.

At STP, 1 mol is 22.4 L, so 23.6 mol would be huge volume, but here compressed to 1L at high T.

P = nRT/V = 23.59375 * 8.314 * 473 / 1

Calculate numerically:

8.314 * 473 = 3932.522
Then * 23.59375 ≈ 3932.522 * 23.59375

First, 3932.522 * 20 = 78,650.44
3932.522 * 3 = 11,797.566
3932.522 * 0.5 = 1,966.261
3932.522 * 0.09375 = ? 3932.522 * 0.09 = 353.92698; 3932.522 * 0.00375 = 14.7469575; total ≈ 368.6739375

Sum: 78,650.44 + 11,797.566 = 90,448.006
+1,966.261 = 92,414.267
+368.674 = 92,782.941 kPa

So approximately 92,783 kPa

But worksheet says: (P)(1) = (755/32)(8.314)(473) → Answer: 92,800 kPa — close enough, rounded.

Good.

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Problem 9: How many moles of O₂ are needed to occupy a volume of 600 mL at 30°C and 100 kPa?

V = 600 mL = 0.6 L
T = 30 + 273 = 303 K
P = 100 kPa
R = 8.314 L·kPa/(mol·K)

n = PV / RT = (100)(0.6) / (8.314)(303)

Numerator: 60
Denominator: 8.314 × 303 ≈ 8.314 × 300 = 2494.2; 8.314 × 3 = 24.942; total 2519.142
n = 60 / 2519.142 ≈ 0.02382 mol

Worksheet says: (100)(0.6) = n(8.314)(303) → Answer: 0.024 mol — matches when rounded.

Good.

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Problem 10: Calculate the volume in m³ of 2 moles of nitrogen at 25°C and 2 atm?

Note: volume in cubic meters.

First, use ideal gas law.

P = 2 atm
n = 2 mol
T = 25 + 273 = 298 K
R = 0.0821 L·atm/(mol·K) — but we need m³.

1 m³ = 1000 L, so better to use R in m³.

R = 0.0821 L·atm/(mol·K) = 0.0821 × 10^{-3} m³·atm/(mol·K) = 8.21 × 10^{-5} m³·atm/(mol·K)

Or use R = 8.314 J/(mol·K) = 8.314 Pa·m³/(mol·K), but P is in atm.

Convert P to Pa: 1 atm = 101325 Pa, so 2 atm = 202650 Pa

R = 8.314 m³·Pa/(mol·K)

V = nRT / P = (2)(8.314)(298) / 202650

Compute numerator: 2 × 8.314 = 16.628; × 298 = 4955.144
Denominator: 202650
V = 4955.144 / 202650 ≈ 0.02445 m³

Using R = 0.0821 L·atm/(mol·K):

V = nRT / P = (2)(0.0821)(298) / 2 = (48.9316) / 2 = 24.4658 L

Convert to m³: 24.4658 / 1000 = 0.0244658 m³ ≈ 0.0245 m³

Worksheet says: (2)(V) = (2)(0.0821)(298) → wait, that would be for P=2, so V = [2*0.0821*298]/2 = 0.0821*298 = 24.4658 L = 0.0245 m³

And it says Answer: 0.0245 m³ — perfect.

Good.

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Now, back to Problem 1 and 3 which had issues.

For Problem 1: We calculated 50.8 L, but worksheet says 46.2 L. Given that other problems match, likely worksheet has error for problem 1.

Similarly, Problem 3 gave 16.4 K, which is correct mathematically but physically odd — probably typo in problem (maybe volume is 4000 mL?).

But since the task is to solve accurately, we provide correct answers.

However, looking at the user's initial request: they uploaded an image with the worksheet, and in the text representation, for problem 1, it shows "(720)(V) = (2)(62.4)(293)" and then "Answer: 46.2 L" — which is arithmetically wrong.

Perhaps they meant T = 273 K? Let's check:

If T = 273 K, V = (2)(62.4)(273)/720 = 34070.4 / 720 = 47.32 L — still not 46.2.

Or if n = 1.8: 1.8*62.4*293/720 = (1.8*62.4=112.32; *293=32909.76; /720=45.7) — close to 46.2.

Or if P = 760: V = (2)(62.4)(293)/760 = 36566.4/760 = 48.11 L — not 46.2.

Another possibility: they used R = 62.4 but T = 20+273=293, but calculated 2*62.4=124.8, 124.8*293.

124.8 * 293:

100*293=29300
24.8*293: 20*293=5860; 4.8*293=1406.4; total 5860+1406.4=7266.4
Sum 29300+7266.4=36566.4 — same.

36566.4 / 720 = let's divide: 720 * 50 = 36000, remainder 566.4, 566.4/720=0.7867, so 50.7867.

Perhaps they did 2*62.4*293 = 2*62.4=124.8, 124.8*293.

Maybe they used 293 as 290: 124.8*290=36192, /720=50.266 — not 46.2.

Or perhaps they used Celsius instead of Kelvin? T=20, then V= (2)(62.4)(20)/720 = 2496/720=3.466 L — no.

I think it's a mistake in the worksheet. For accuracy, we'll use 50.8 L.

But to match the format, perhaps the expected answer is 46.2, but that would be incorrect.

Let's look at problem 3 similarly.

Problem 3: T = PV/(nR) = (800*0.4)/( (10/32)*62.4 ) = 320 / (0.3125*62.4) = 320 / 19.5 = 16.41 K

If volume was 4.0 L, then V=4.0, P=800, n=0.3125, R=62.4, T= (800*4.0)/(0.3125*62.4) = 3200 / 19.5 = 164.1 K — more reasonable.

Probably typo in problem: volume is 4000 mL = 4.0 L, not 400 mL.

In many textbooks, such typos occur.

Given that, and since the worksheet shows 16.4 K for 400 mL, we'll keep it as is for now.

But for the sake of this response, I'll provide the mathematically correct answers based on given numbers.

So summarizing:

1. 50.8 L
2. 27.8 atm
3. 16.4 K
4. 0.86 mol
5. 11.6 L
6. 18.8 atm
7. 122 K
8. 92,800 kPa
9. 0.024 mol
10. 0.0245 m³

But for problem 1, let's see if there's another interpretation.

Perhaps "nitrogen" is N2, molar mass 28, but no, moles are given as 2, so mass not needed.

Another thought: in problem 1, if they used R = 0.0821 and P in atm, but forgot to convert 720 mmHg to atm, and used P=720, then V = (2)(0.0821)(293)/720 = 48.1106/720 = 0.0668 L — no.

Or if they used T in Celsius: V = (2)(0.0821)(20)/0.9474 = 3.284/0.9474 = 3.466 L — no.

I think we have to go with 50.8 L.

Perhaps the answer 46.2 is for a different set of numbers.

Let's calculate what would give 46.2 L.

From V = nRT/P = 2 * 62.4 * T / 720 = 46.2

So 124.8 * T / 720 = 46.2
124.8 T = 46.2 * 720 = 33264
T = 33264 / 124.8 = 266.5 K = -6.5°C — not 20°C.

So not matching.

Therefore, for accuracy, I'll box the correct answers.

Final Answers:

1. 50.8 L
2. 27.8 atm
3. 16.4 K
4. 0.86 mol
5. 11.6 L
6. 18.8 atm
7. 122 K
8. 92,800 kPa
9. 0.024 mol
10. 0.0245 m³

But to match the worksheet's style, perhaps round appropriately.

For problem 1, since worksheet says 46.2, but it's wrong, I'll put the correct one.

Perhaps in the context, they expect us to use the formula as written, but the arithmetic is off.

Another idea: in problem 1, they have "(720)(V) = (2)(62.4)(293)" and then say Answer: 46.2 L — but 2*62.4*293 = 36,566.4, 36,566.4 / 720 = 50.7867, not 46.2.

Unless they meant (720)(V) = (2)(62.4)(273) — then 2*62.4*273 = 34,070.4, /720 = 47.32 — still not 46.2.

Or if n=1.8: 1.8*62.4*293=32,909.76, /720=45.7 — close to 46.2.

Perhaps it's 2 moles but they used molar mass or something.

I think for the purpose of this, since the instruction is to solve accurately, I'll provide the correct calculation.

So for problem 1: 50.8 L

But let's list all final answers as per accurate calculation.

Final Answer:
1. 50.8 L
2. 27.8 atm
3. 16.4 K
4. 0.86 mol
5. 11.6 L
6. 18.8 atm
7. 122 K
8. 92,800 kPa
9. 0.024 mol
10. 0.0245 m³

To make it neat, and since the worksheet might have specific rounding, but I think this is fine.

For problem 3, 16.4 K is correct based on given numbers, even if unrealistic.

So I'll go with that.