Combined Gas Law Problems worksheet with four questions on pressure, volume, and temperature changes.

Let's solve each of these Combined Gas Law problems step by step.

The Combined Gas Law is:

$$
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
$$

Where:
- $ P $ = pressure (in atm)
- $ V $ = volume (in liters)
- $ T $ = temperature (in Kelvin, K)

Remember: Convert Celsius to Kelvin using $ T(K) = T(°C) + 273 $

---

Problem 1

Given:
- $ P_1 = 12 \text{ atm} $
- $ V_1 = 23 \text{ L} $
- $ T_1 = 200 \text{ K} $
- $ P_2 = 14 \text{ atm} $
- $ T_2 = 300 \text{ K} $
- Find: $ V_2 = ? $

Use the combined gas law:

$$
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
$$

Solve for $ V_2 $:

$$
V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}
$$

Plug in values:

$$
V_2 = 23 \times \frac{12}{14} \times \frac{300}{200}
$$

$$
V_2 = 23 \times 0.8571 \times 1.5
$$

$$
V_2 = 23 \times 1.2857 \approx 29.57 \text{ L}
$$

✔ Answer: $ \boxed{29.6} $ liters (rounded to one decimal place)

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Problem 2

Given:
- $ V_1 = 17 \text{ L} $
- $ P_1 = 2.3 \text{ atm} $
- $ T_1 = 299 \text{ K} $
- $ T_2 = 350 \text{ K} $
- $ P_2 = 1.5 \text{ atm} $
- Find: $ V_2 = ? $

Use:

$$
V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}
$$

$$
V_2 = 17 \times \frac{2.3}{1.5} \times \frac{350}{299}
$$

First, compute:

- $ \frac{2.3}{1.5} = 1.5333 $
- $ \frac{350}{299} \approx 1.1706 $

Now:

$$
V_2 = 17 \times 1.5333 \times 1.1706 \approx 17 \times 1.794 \approx 30.498 \text{ L}
$$

✔ Answer: $ \boxed{30.5} $ liters (rounded to one decimal place)

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Problem 3

Given:
- $ V_1 = 28 \text{ L} $
- $ T_1 = 45^\circ C = 45 + 273 = 318 \text{ K} $
- $ V_2 = 34 \text{ L} $
- $ T_2 = 35^\circ C = 35 + 273 = 308 \text{ K} $
- $ P_2 = 2.0 \text{ atm} $
- Find: $ P_1 = ? $

Use:

$$
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
$$

Solve for $ P_1 $:

$$
P_1 = P_2 \times \frac{V_2}{V_1} \times \frac{T_1}{T_2}
$$

$$
P_1 = 2.0 \times \frac{34}{28} \times \frac{318}{308}
$$

Compute:

- $ \frac{34}{28} = 1.2143 $
- $ \frac{318}{308} \approx 1.0325 $

$$
P_1 = 2.0 \times 1.2143 \times 1.0325 \approx 2.0 \times 1.252 \approx 2.504 \text{ atm}
$$

✔ Answer: $ \boxed{2.5} $ atm (rounded to one decimal place)

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Problem 4

Given:
- $ T_1 = 14^\circ C = 14 + 273 = 287 \text{ K} $
- $ V_1 = 4.5 \text{ L} $
- $ T_2 = 29^\circ C = 29 + 273 = 302 \text{ K} $
- Pressure is constant → Use Charles’s Law: $ \frac{V_1}{T_1} = \frac{V_2}{T_2} $

So:

$$
V_2 = V_1 \times \frac{T_2}{T_1} = 4.5 \times \frac{302}{287}
$$

$$
\frac{302}{287} \approx 1.0523
$$

$$
V_2 = 4.5 \times 1.0523 \approx 4.735 \text{ L}
$$

✔ Answer: $ \boxed{4.7} $ liters (rounded to one decimal place)

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✔ Final Answers Summary:

1) $ \boxed{29.6} $ L
2) $ \boxed{30.5} $ L
3) $ \boxed{2.5} $ atm
4) $ \boxed{4.7} $ L

Let me know if you'd like a visual explanation or help with units!