Worksheet: Combined Gas Law and Ideal Gas Law ... - Free Printable

Let's solve each problem step by step using the Combined Gas Law and Ideal Gas Law.

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Problem 1:

> A 952 cm³ container of gas is exerting a pressure of 108 kPa while at a temperature of 48°C. Calculate the pressure of this same amount of gas in a 1236 cm³ container at a temperature of 64°C.

#### Step 1: Identify knowns
- $ V_1 = 952 \text{ cm}^3 $
- $ P_1 = 108 \text{ kPa} $
- $ T_1 = 48^\circ\text{C} = 48 + 273 = 321 \text{ K} $
- $ V_2 = 1236 \text{ cm}^3 $
- $ T_2 = 64^\circ\text{C} = 64 + 273 = 337 \text{ K} $
- $ P_2 = ? $

Since the amount of gas (moles) is constant, use the Combined Gas Law:

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

Solve for $ P_2 $:

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

Plug in values:

$$
P_2 = 108 \cdot \frac{952}{1236} \cdot \frac{337}{321}
$$

Calculate step-by-step:

- $ \frac{952}{1236} \approx 0.7698 $
- $ \frac{337}{321} \approx 1.050 $
- $ P_2 = 108 \cdot 0.7698 \cdot 1.050 \approx 108 \cdot 0.8083 \approx 87.3 \text{ kPa} $

✔ Answer: $ \boxed{87.3} $ kPa

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Problem 2:

> At STP, a sample of gas occupies 24.5 mL. Calculate the volume of this gas at a pressure of 2.3 atm and a temperature of 301 K.

#### Step 1: Understand STP conditions
STP = Standard Temperature and Pressure:
- $ T_1 = 0^\circ\text{C} = 273 \text{ K} $
- $ P_1 = 1 \text{ atm} $
- $ V_1 = 24.5 \text{ mL} $
- $ P_2 = 2.3 \text{ atm} $
- $ T_2 = 301 \text{ K} $
- $ V_2 = ? $

Use Combined Gas Law:

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

Plug in:

$$
V_2 = 24.5 \cdot \frac{1}{2.3} \cdot \frac{301}{273}
$$

Calculate:

- $ \frac{1}{2.3} \approx 0.4348 $
- $ \frac{301}{273} \approx 1.1026 $
- $ V_2 = 24.5 \cdot 0.4348 \cdot 1.1026 \approx 24.5 \cdot 0.4794 \approx 11.70 \text{ mL} $

✔ Answer: $ \boxed{11.7} $ mL

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Problem 3:

> A 3.25 L container of ammonia gas exerts a pressure of 652 mm Hg at a temperature of 243 K. Calculate the pressure of this same amount of gas in a 2.50 L container at a temperature of 221 K.

#### Knowns:
- $ V_1 = 3.25 \text{ L} $
- $ P_1 = 652 \text{ mm Hg} $
- $ T_1 = 243 \text{ K} $
- $ V_2 = 2.50 \text{ L} $
- $ T_2 = 221 \text{ K} $
- $ P_2 = ? $

Use Combined Gas Law:

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

Wait! Note: Volume decreases (3.25 → 2.50), so pressure increases. But temperature decreases (243 → 221), so pressure decreases. We'll see which dominates.

$$
P_2 = 652 \cdot \frac{3.25}{2.50} \cdot \frac{221}{243}
$$

Calculate:

- $ \frac{3.25}{2.50} = 1.3 $
- $ \frac{221}{243} \approx 0.9095 $
- $ P_2 = 652 \cdot 1.3 \cdot 0.9095 \approx 652 \cdot 1.18235 \approx 770.5 \text{ mm Hg} $

✔ Answer: $ \boxed{771} $ mm Hg (rounded to nearest whole number)

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Problem 4:

> A sample of gas has a volume of 5.23 cm³ at a pressure of 72.6 kPa and a temperature of 25°C. What will be the volume of the gas if the pressure is changed to 124 kPa and the temperature is changed to 0°C?

#### Knowns:
- $ V_1 = 5.23 \text{ cm}^3 $
- $ P_1 = 72.6 \text{ kPa} $
- $ T_1 = 25^\circ\text{C} = 298 \text{ K} $
- $ P_2 = 124 \text{ kPa} $
- $ T_2 = 0^\circ\text{C} = 273 \text{ K} $
- $ V_2 = ? $

Use Combined Gas Law:

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

$$
V_2 = 5.23 \cdot \frac{72.6}{124} \cdot \frac{273}{298}
$$

Calculate:

- $ \frac{72.6}{124} \approx 0.5855 $
- $ \frac{273}{298} \approx 0.9161 $
- $ V_2 = 5.23 \cdot 0.5855 \cdot 0.9161 \approx 5.23 \cdot 0.5352 \approx 2.80 \text{ cm}^3 $

✔ Answer: $ \boxed{2.80} $ cm³

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✔ Final Answers:

1. 87.3 kPa
2. 11.7 mL
3. 771 mm Hg
4. 2.80 cm³

Let me know if you'd like these solved using the Ideal Gas Law instead (though Combined Gas Law is sufficient since moles are constant).