Ideal Gas Law Worksheet with step-by-step solutions and handwritten answers using the PV = nRT formula.

Let’s solve problem #4 step by step.

We are given:
- Moles of gas (n) = 7.7 mol
- Pressure (P) = 0.09 atm
- Temperature (T) = 56 °C

We need to find the volume (V).

First, remember the Ideal Gas Law:
PV = nRT

We want to solve for V, so rearrange the formula:
V = nRT / P

But wait — temperature must be in Kelvin, not Celsius!
So convert 56 °C to Kelvin:
T(K) = 56 + 273 = 329 K

Now plug in the values:
- n = 7.7 mol
- R = 0.0821 L·atm/(K·mol) ← we use this because pressure is in atm
- T = 329 K
- P = 0.09 atm

Calculate numerator first:
n × R × T = 7.7 × 0.0821 × 329

Step-by-step multiplication:

First, 7.7 × 0.0821 = let’s calculate that:
7.7 × 0.08 = 0.616
7.7 × 0.0021 = 0.01617
Add them: 0.616 + 0.01617 = 0.63217

Now multiply by 329:
0.63217 × 329

Break it down:
0.63217 × 300 = 189.651
0.63217 × 29 = let’s do 0.63217 × 30 = 18.9651, minus 0.63217 = 18.33293
Total = 189.651 + 18.33293 = 207.98393

So numerator ≈ 207.98

Now divide by P = 0.09:
V = 207.98 / 0.09 = ?

Dividing by 0.09 is same as multiplying by 100/9 ≈ 11.111...

207.98 ÷ 0.09 = 207.98 × (100/9) = 20798 / 9 = let’s compute:

9 × 2310 = 20790 → remainder 8 → so 2310 + 8/9 ≈ 2310.89

Wait — that seems too big. Let me double-check my math.

Actually, let’s recalculate more carefully:

nRT = 7.7 × 0.0821 × 329

Do 0.0821 × 329 first:

0.0821 × 300 = 24.63
0.0821 × 29 = 2.3809
Total = 24.63 + 2.3809 = 27.0109

Now multiply by 7.7:

27.0109 × 7 = 189.0763
27.0109 × 0.7 = 18.90763
Total = 189.0763 + 18.90763 = 207.98393 ← same as before

Now divide by 0.09:

207.98393 ÷ 0.09 = ?

Move decimal: 20798.393 ÷ 9 = ?

9 × 2310 = 20790 → subtract → 8.393 left
8.393 ÷ 9 ≈ 0.93255...

So total V ≈ 2310.93 liters

That still feels huge — but let’s think: very low pressure (0.09 atm), lots of moles (7.7), and warm temp — so yes, volume should be large.

Alternatively, maybe I made a mistake in unit? No — all units match: R is in L·atm/(K·mol), P in atm, T in K, n in mol → V comes out in liters.

Let me verify with calculator-style steps:

7.7 × 0.0821 = 0.63217
0.63217 × 329 = let’s do 0.63217 × 330 = 208.6161, minus 0.63217 = 207.98393 → correct

207.98393 ÷ 0.09 = 2310.93255...

So approximately 2311 liters

But let’s round appropriately. The inputs have 2 significant figures in pressure (0.09 has one sig fig? Wait — 0.09 could be considered 1 sig fig, but often in these problems, if written as 0.09, it might imply two — actually, leading zeros don’t count, so 0.09 has one sig fig. But 7.7 has two, 56 has two — so probably answer should have two sig figs.

Wait — 0.09 atm — is that one or two sig figs? In many textbooks, if it's written as 0.09, it's ambiguous, but often treated as one. However, looking at other problems on the worksheet, they keep several decimals. Since the problem didn't specify, and others did precise answers, let’s go with what we calculated.

But let’s check if I used correct R — yes, 0.0821 for atm.

Another way: perhaps the temperature conversion? 56 + 273 = 329 — correct.

Maybe I can write it as:

V = (7.7 * 0.0821 * 329) / 0.09

Compute inside: 7.7 * 329 = 2533.3
Then 2533.3 * 0.0821 = let’s see: 2500*0.0821=205.25, 33.3*0.0821≈2.734, total ≈207.984 — same

Then /0.09 = 2310.93

So final answer is about 2311 liters.

But let’s see if the worksheet expects rounding. Problem 1 had T=204.63 K, problem 2 had n=1.2586 mol, problem 3 had P=1.642 atm — so they kept several decimals.

So I’ll report as 2310.93 L, but perhaps round to nearest whole number since volume in liters usually doesn’t need decimals unless specified.

Wait — 2310.93 is very close to 2311.

But let me do exact calculation:

7.7 × 0.0821 = 0.63217
0.63217 × 329 = 0.63217 × 300 = 189.651; 0.63217 × 29 = 18.33293; sum 207.98393
207.98393 / 0.09 = 2310.932555...

So 2310.93 L if keeping two decimals, or 2311 L if rounded.

Looking at problem 3, they gave both atm and kPa, but here no conversion needed.

I think for consistency, since other answers had up to 4 decimals, but volume might be fine as whole number.

Actually, let’s see the input: 7.7 moles (two sig figs), 0.09 atm (one sig fig?), 56°C (two sig figs). The limiting is 0.09 which has one sig fig — so answer should have one sig fig? That would be 2000 L — but that seems too rough.

In educational contexts, often they treat 0.09 as having two sig figs if it’s from measurement. To be safe, I’ll provide the calculated value and note.

But since the worksheet shows precise answers, I’ll go with 2311 L.

Wait — I recall that in some systems, trailing zeros after decimal matter, but here 0.09 — the '9' is the only significant digit, so one sig fig. But that would make answer 2000 L, which is odd.

Perhaps it's 0.090? But it's written as 0.09.

Looking back at the image, problem 4 says "pressure of 0.09 atm" — likely intended as two sig figs, as 9 hundredths.

In many textbooks, 0.09 is considered to have one sig fig, but in practice for such problems, they expect the calculation as is.

To match the style of the worksheet, I'll report the answer as 2311 L.

But let me calculate once more with exact fractions.

R = 0.0821 exactly as given.

n = 7.7 = 77/10
T = 329
P = 0.09 = 9/100

V = (77/10 * 0.0821 * 329) / (9/100) = (77/10 * 0.0821 * 329) * (100/9)

= 77 * 0.0821 * 329 * 100 / (10 * 9) = 77 * 0.0821 * 329 * 10 / 9

Compute step by step:

First, 77 * 10 = 770

Then 770 / 9 = 85.555...

Better: 77 * 329 = 77*300=23100, 77*29=2233, total 25333

Then 25333 * 0.0821 = ?

25333 * 0.08 = 2026.64
25333 * 0.002 = 50.666
25333 * 0.0001 = 2.5333
Sum: 2026.64 + 50.666 = 2077.306; +2.5333=2079.8393? Wait no:

0.0821 = 0.08 + 0.002 + 0.0001

But 25333 * 0.08 = 2026.64
25333 * 0.002 = 50.666
25333 * 0.0001 = 2.5333
Total = 2026.64 + 50.666 = 2077.306; +2.5333 = 2079.8393 — but this is for 25333 * 0.0821, but earlier we had 7.7 * 0.0821 * 329 = 207.98393, and 7.7*329=2533.3, not 25333. I missed a decimal.

7.7 * 329 = let's calculate: 7*329=2303, 0.7*329=230.3, total 2533.3 — yes.

So 2533.3 * 0.0821

2533.3 * 0.08 = 202.664
2533.3 * 0.002 = 5.0666
2533.3 * 0.0001 = 0.25333
Sum: 202.664 + 5.0666 = 207.7306; +0.25333 = 207.98393 — correct.

Then /0.09 = 207.98393 / 0.09 = 2310.93255...

So V = 2310.93 L

I think it's safe to box 2311 L as the final answer, rounding to nearest liter.

Or perhaps keep one decimal? But others didn't.

Problem 1: T=204.63 K — two decimals
Problem 2: n=1.2586 mol — four decimals
Problem 3: P=1.642 atm — three decimals

So for volume, 2310.93 L is fine, but since it's large, maybe 2311 L.

I'll go with 2311 L.

Final Answer:
2311 L