Worksheet on solving ideal gas law problems with temperature, pressure, volume, and moles.

It looks like you've uploaded a worksheet titled "Ideal Gas Law" with four problems involving the Ideal Gas Law, which is expressed as:

$$
PV = nRT
$$

Where:
- $P$ = pressure (in atm)
- $V$ = volume (in liters, L)
- $n$ = number of moles
- $R$ = ideal gas constant = $0.0821 \frac{\text{L·atm}}{\text{mol·K}}$
- $T$ = temperature (in Kelvin, K)

Let’s solve each problem step by step.

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Problem 1:

> How many moles of gas are required to occupy a pressure of 2.5 atmospheres and a temperature of 300 K?

Wait — this problem seems incomplete. It mentions pressure (2.5 atm) and temperature (300 K), but no volume is given. Without volume or another variable, we cannot solve for moles.

But looking at the text again:
> "How many moles of gas are required to occupy a pressure of 2.5 atmospheres and a temperature of 300 K?"

This appears to be missing the volume. Unless there's a typo or omission, this problem cannot be solved as stated.

✔ Conclusion: Problem 1 is missing information (volume). Please check if the volume was supposed to be included (e.g., maybe it was meant to say “a 10 L container” or something similar).

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Problem 2:

> If I have a 25 L container that holds 2.0 moles of gas at a pressure of 345 atm, what is the temperature in units of the container?

We use the Ideal Gas Law:

$$
PV = nRT
\Rightarrow T = \frac{PV}{nR}
$$

Given:
- $P = 345$ atm
- $V = 25$ L
- $n = 2.0$ mol
- $R = 0.0821 \frac{\text{L·atm}}{\text{mol·K}}$

Plug in:

$$
T = \frac{(345)(25)}{(2.0)(0.0821)} = \frac{8625}{0.1642} \approx 52,534 \text{ K}
$$

That’s extremely high — about 52,500 K, which is hotter than the surface of the Sun! This suggests either:
- A typo in the pressure (345 atm is very high),
- Or perhaps the pressure is 3.45 atm instead of 345 atm.

Let’s double-check: if $P = 3.45$ atm:

$$
T = \frac{(3.45)(25)}{(2.0)(0.0821)} = \frac{86.25}{0.1642} \approx 525.3 \text{ K}
$$

That’s more reasonable (~252°C).

✔ Likely typo: Pressure should be 3.45 atm, not 345 atm.

But based on the given values:

$$
\boxed{T \approx 52,534 \text{ K}}
$$

(But this is physically unrealistic unless it's a special context.)

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Problem 3:

> It is not safe to put aerosol cans in a car trunk, because the pressure inside the can increases when heated. If a 1.0 L canister initially has 1.2 moles of gas and the temperature of the canister is 300 degrees Celsius, what is the pressure inside the canister?

First, convert temperature to Kelvin:

$$
T = 300^\circ C + 273 = 573 \text{ K}
$$

Given:
- $V = 1.0$ L
- $n = 1.2$ mol
- $T = 573$ K
- $R = 0.0821$

Use:

$$
P = \frac{nRT}{V} = \frac{(1.2)(0.0821)(573)}{1.0}
$$

Calculate step-by-step:

- $1.2 \times 0.0821 = 0.09852$
- $0.09852 \times 573 \approx 56.36$

So:

$$
P \approx 56.4 \text{ atm}
$$

✔ Answer: The pressure inside the canister is approximately 56.4 atm.

This is very high — explains why aerosol cans can explode when heated.

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Problem 4:

> How many moles of gas are in a 10 L cylinder if the temperature is 300 K and the pressure is 200 atm?

Use:

$$
n = \frac{PV}{RT}
$$

Given:
- $P = 200$ atm
- $V = 10$ L
- $T = 300$ K
- $R = 0.0821$

$$
n = \frac{(200)(10)}{(0.0821)(300)} = \frac{2000}{24.63} \approx 81.2 \text{ moles}
$$

✔ Answer: Approximately 81.2 moles of gas.

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✔ Final Answers:

1. Incomplete – missing volume. Cannot solve.
2. $T \approx 52,534 \text{ K}$ (but likely a typo; check pressure).
3. $P \approx 56.4 \text{ atm}$
4. $n \approx 81.2 \text{ moles}$

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Let me know if you'd like help correcting any typos or if you want these problems rewritten with realistic values!