Completed answer key for an Ideal Gas Law worksheet showing how to solve for variables like temperature and moles.

Let’s solve problem #4 step by step.

We are told:
- n = 7.7 moles
- P = 0.09 atm
- T = 56 °C

We need to find the volume (V) using the ideal gas law:
PV = nRT

But first — temperature must be in Kelvin, not Celsius!

Step 1: Convert °C to K

T(K) = T(°C) + 273
T = 56 + 273 = 329 K

Step 2: Rearrange PV = nRT to solve for V

V = nRT / P

Step 3: Plug in the values

R = 0.0821 L·atm/(K·mol) ← we use this because pressure is in atm

V = (7.7 mol × 0.0821 L·atm/(K·mol) × 329 K) / 0.09 atm

Let’s calculate step by step:

First, multiply 7.7 × 0.0821 =
7.7 × 0.0821 = let’s do it carefully:
7 × 0.0821 = 0.5747
0.7 × 0.0821 = 0.05747
Total = 0.5747 + 0.05747 = 0.63217

Now multiply that by 329:
0.63217 × 329

Break it down:
0.63217 × 300 = 189.651
0.63217 × 29 = let’s compute:
0.63217 × 30 = 18.9651 → minus 0.63217 = 18.33293
So total = 189.651 + 18.33293 = 207.98393

Now divide by 0.09:
207.98393 ÷ 0.09 = ?

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

207.98393 × (100/9) = 20798.393 / 9

Do the division:
9 × 2310 = 20790
Remainder 8.393 → so about 2310 + 0.932 ≈ 2310.932

Wait — that can’t be right. Let me recalculate more accurately.

Actually, let’s redo the multiplication with better precision.

Use calculator-style steps:

nRT = 7.7 × 0.0821 × 329

First: 0.0821 × 329 = ?

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

Now multiply by 7.7:
27.0109 × 7.7

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

Now divide by P = 0.09:

207.98393 ÷ 0.09 = 207.98393 ÷ (9/100) = 207.98393 × (100/9)

= 20798.393 ÷ 9

Now divide:

9 × 2310 = 20790
Subtract: 20798.393 - 20790 = 8.393
8.393 ÷ 9 = 0.93255...

So total V = 2310 + 0.93255 ≈ 2310.93 L

That seems huge — but let’s check if units make sense.

7.7 moles at low pressure (0.09 atm) and moderate temp (329 K) — yes, volume should be large.

For comparison: at STP (1 atm, 273 K), 1 mole = 22.4 L → 7.7 moles would be ~172 L. But here pressure is only 0.09 atm (about 1/11th of 1 atm), so volume should be about 11 times larger → 172 × 11 ≈ 1892 L — close to our answer. And temperature is higher too (329 vs 273), so even bigger. So 2310 L makes sense.

But let’s double-check calculation once more:

nRT/P = (7.7 * 0.0821 * 329) / 0.09

Compute numerator: 7.7 * 0.0821 = 0.63217
0.63217 * 329 = let's do exact:

0.63217 * 300 = 189.651
0.63217 * 29 = 18.33293
Sum: 207.98393 — correct

207.98393 / 0.09 = 2310.93255...

So approximately 2310.93 liters

But maybe round to reasonable sig figs?

Given values:
- 7.7 moles → 2 sig figs
- 0.09 atm → 1 sig fig? Wait, 0.09 could be considered 1 or 2? Usually leading zeros don't count, so 1 sig fig.
- 56°C → 2 sig figs

The limiting is probably 0.09 atm (1 sig fig) — but that would give answer with 1 sig fig → 2000 L? That seems too rough.

In many school contexts, they expect you to keep a few decimals unless specified.

Looking back at other problems on worksheet — they kept 2-4 decimal places.

Problem 1: T=204.63 K
Problem 2: n=1.2586 mol
Problem 3: P=1.642 atm or 166.28 kPa

So likely, we should report with similar precision.

Our value is 2310.93 L — but perhaps write as 2311 L or keep one decimal?

Wait — let’s see what the exact calculation gives.

Actually, let me use more precise multiplication:

7.7 × 0.0821 = 7.7 × 0.0821

7.7 × 0.08 = 0.616
7.7 × 0.0021 = 0.01617
Total = 0.63217 — correct

0.63217 × 329:

Let me compute 63217 × 329 / 100000

Better: 0.63217 × 329 = ?

329 × 0.6 = 197.4
329 × 0.03 = 9.87
329 × 0.002 = 0.658
329 × 0.00017 = 0.05593
Add them:

197.4 + 9.87 = 207.27
+0.658 = 207.928
+0.05593 = 207.98393 — same

Then /0.09 = 207.98393 / 0.09

= 2310.932555...

So, to match the style of the worksheet, I’ll round to two decimal places: 2310.93 L

But wait — in problem 3, they gave both atm and kPa. Here, no conversion needed since question doesn’t ask for different unit.

Also, note: in problem 4, the image has blurred work, but we’re solving from scratch.

Final Answer: Volume is approximately 2310.93 liters.

But let me confirm with direct formula:

V = nRT/P = (7.7)(0.0821)(329)/(0.09)

Calculate numerically:

First, 0.0821 * 329 = 27.0109

Then 7.7 * 27.0109 = 207.98393

Then 207.98393 / 0.09 = 2310.932555...

Yes.

So, final answer is 2310.93 L (rounded to two decimals).

But perhaps the worksheet expects whole number? Looking at problem 1: 204.63 — so decimals are fine.

I think 2310.93 is acceptable.

However, let’s consider significant figures again.

Input values:
- 7.7 → 2 sig figs
- 0.09 → ambiguous, but often treated as 1 sig fig (since leading zero doesn’t count)
- 56 → 2 sig figs

The constant R has 3 sig figs (0.0821)

The least precise measurement is 0.09 atm — which might be 1 sig fig. If so, answer should have 1 sig fig → 2000 L.

But that seems too crude, and in educational contexts, sometimes they ignore that for intermediate steps.

Looking at problem 2: pressure 1.2 atm (2 sig figs), volume 31 L (2 sig figs), temp 87°C (2 sig figs) → answer given as 1.2586 mol — which is 5 sig figs! So clearly, they are not strict about sig figs here.

Therefore, I will report the calculated value as is.

Final Answer: 2310.93 L

But to match the format of other answers (like 204.63, 1.2586, etc.), I'll write it as:

2310.93

Units are liters, as per context.

In problem 4, the question says “what is the volume?” — and since inputs are in atm and liters, output is in liters.

So, boxed answer should be 2310.93

But let me write it properly.

Actually, in the image, for problem 3, they wrote P=1.642 atm — three decimals.

So for consistency, I’ll keep two decimals for volume.

2310.93 is fine.

Perhaps they expect rounding to nearest whole number? But 2310.93 is very close to 2311.

But I’ll stick with 2310.93 as per calculation.

Wait — let me do one last check with calculator simulation:

7.7 * 0.0821 = 0.63217

0.63217 * 329 = let's compute:

329 * 0.63217

300*0.63217=189.651

29*0.63217=18.33293

Sum 207.98393

207.98393 / 0.09 = 2310.932555...

Yes.

So, Final Answer: 2310.93

But in the context, perhaps write as 2311 if rounded, but I think 2310.93 is accurate.

Looking back at problem 1: they have T=204.63 K — which is to two decimals.

Similarly, here, volume to two decimals is fine.

So, I'll go with that.

Final Answer: 2310.93