Combined Gas Law Worksheet - Solutions showing step-by-step calculations for gas volume changes under varying pressure and temperature conditions.

Here are the step-by-step solutions for the problems shown in the worksheet.

Problem 1

Question: If I initially have 4.0 L of a gas at a pressure of 1.1 atm, what will the volume be if I increase the pressure to 3.4 atm?

Step-by-Step Solution:
1. Identify the variables:
* Initial Pressure ($P_1$) = $1.1 \text{ atm}$
* Initial Volume ($V_1$) = $4.0 \text{ L}$
* Final Pressure ($P_2$) = $3.4 \text{ atm}$
* Final Volume ($V_2$) = ?
2. Choose the formula: Since temperature is not mentioned, we assume it stays constant. We use Boyle's Law:
$$P_1 V_1 = P_2 V_2$$
3. Plug in the numbers:
$$(1.1)(4.0) = (3.4)(V_2)$$
4. Solve for $V_2$:
$$4.4 = 3.4 V_2$$
$$V_2 = \frac{4.4}{3.4}$$
$$V_2 \approx 1.29 \text{ L}$$

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Problem 2

Question: A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is $20^\circ\text{C}$, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is $-15^\circ\text{C}$?

Step-by-Step Solution:
1. Convert Temperatures to Kelvin: Gas law calculations must always use Kelvin.
* $T_1 = 20 + 273 = 293 \text{ K}$
* $T_2 = -15 + 273 = 258 \text{ K}$
2. Identify the other variables:
* $P_1 = 1.05 \text{ atm}$
* $V_1 = 5.0 \text{ L}$
* $P_2 = 0.65 \text{ atm}$
* $V_2 = ?$
3. Choose the formula: Use the Combined Gas Law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
4. Plug in the numbers:
$$\frac{(1.05)(5.0)}{293} = \frac{(0.65)(V_2)}{258}$$
5. Solve for $V_2$:
$$\frac{5.25}{293} = \frac{0.65 V_2}{258}$$
$$0.0179 = \frac{0.65 V_2}{258}$$
Multiply both sides by 258:
$$4.62 = 0.65 V_2$$
Divide by 0.65:
$$V_2 = \frac{4.62}{0.65}$$
$$V_2 \approx 7.11 \text{ L}$$

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Problem 3

Question: A small research submarine with a volume of $1.2 \times 10^5 \text{ L}$ has an internal pressure of 1.0 atm and an internal temperature of $15^\circ\text{C}$. If the submarine descends to a depth where the pressure is 150 atm and the temperature is $3^\circ\text{C}$, what will the volume of the gas inside be if the hull of the submarine breaks?

Step-by-Step Solution:
1. Convert Temperatures to Kelvin:
* $T_1 = 15 + 273 = 288 \text{ K}$
* $T_2 = 3 + 273 = 276 \text{ K}$
2. Identify the other variables:
* $P_1 = 1.0 \text{ atm}$
* $V_1 = 1.2 \times 10^5 \text{ L} = 120,000 \text{ L}$
* $P_2 = 150 \text{ atm}$
* $V_2 = ?$
3. Choose the formula: Combined Gas Law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
4. Plug in the numbers:
$$\frac{(1.0)(120,000)}{288} = \frac{(150)(V_2)}{276}$$
5. Solve for $V_2$:
Left side: $\frac{120,000}{288} \approx 416.67$
Equation becomes: $416.67 = \frac{150 V_2}{276}$
Multiply by 276:
$$115,000 = 150 V_2$$
Divide by 150:
$$V_2 = \frac{115,000}{150}$$
$$V_2 \approx 767 \text{ L}$$

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Problem 4

Question: People who are angry sometimes say that they feel as if they'll explode. If a calm person with a lung capacity of 3.5 liters and a body temperature of $36^\circ\text{C}$ gets angry, what will the volume of the person's lungs be if their temperature rises to $39^\circ\text{C}$. Based on this, do you think it's likely they will explode?

Step-by-Step Solution:
1. Convert Temperatures to Kelvin:
* $T_1 = 36 + 273 = 309 \text{ K}$
* $T_2 = 39 + 273 = 312 \text{ K}$
2. Identify the other variables:
* $V_1 = 3.5 \text{ L}$
* $V_2 = ?$
* (Pressure is assumed constant inside the body relative to the outside).
3. Choose the formula: Charles's Law (Volume and Temperature relationship):
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
4. Plug in the numbers:
$$\frac{3.5}{309} = \frac{V_2}{312}$$
5. Solve for $V_2$:
$$0.01132 = \frac{V_2}{312}$$
$$V_2 = 0.01132 \times 312$$
$$V_2 \approx 3.53 \text{ L}$$

Conclusion: The volume only increased from 3.5 L to 3.53 L. This is a very tiny change, so it is unlikely they would explode.

Final Answer:
1) 1.29 L
2) 7.11 L
3) 767 L
4) 3.53 L