Conceptual Chemistry Practice Page: Gas Laws and Weather Balloon Behavior

Let's solve each part of this Conceptual Chemistry Practice Page step by step.

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1. Principle Difference Between a Liquid and a Gas Under Pressure

We are asked to fill in the blanks based on how liquids and gases behave under pressure.

#### For a liquid under pressure:
- Its volume: [doesn't change noticeably]
- Reason: Liquids are nearly incompressible. The molecules are already close together, so applying pressure doesn’t significantly reduce their volume.
- Its density: [doesn't change noticeably]
- Since volume doesn’t change much and mass stays constant, density remains nearly the same.

#### For a gas under pressure:
- Its volume: [decreases]
- Gases are compressible. When pressure increases, gas particles are pushed closer together, reducing volume.
- Its density: [increases]
- Density = mass/volume. If mass stays the same but volume decreases, density increases.

✔ Answers:
- Liquid under pressure:
- Volume: doesn't change noticeably
- Density: doesn't change noticeably
- Gas under pressure:
- Volume: decreases
- Density: increases

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2. Sketch and Explanation of Weather Balloon at High Altitude

#### Sketch Instructions:
You're told to draw the weather balloon at high altitude.
Since the balloon is rising into the atmosphere where pressure is lower, the air inside the balloon will expand.

So, your sketch should show:
- A larger balloon than at ground level.
- Possibly with more stretched or expanded shape (but still intact — it may eventually burst).

#### Explanation:
At higher altitudes, atmospheric pressure decreases. The pressure outside the balloon becomes much lower than the pressure inside. This causes the gas inside the balloon to expand, increasing its volume. Because the balloon is elastic (like latex), it stretches to accommodate the larger volume.

➡️ So, the balloon gets bigger as it rises because the external pressure decreases, allowing the internal gas to expand.

✔ Answer:
> At high altitude, the weather balloon is larger than at ground level. This happens because atmospheric pressure decreases with altitude, so the gas inside the balloon expands to fill more space due to reduced external pressure.

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3. Using Boyle’s Law to Calculate New Volume

Boyle’s Law states:
\[
P_1 V_1 = P_2 V_2
\]

Given:
- \( P_1 = 1 \text{ atm} \) (at ground level)
- \( V_1 = 10 \text{ m}^3 \)
- \( P_2 = 0.3 \text{ atm} \) (at 10 km altitude)
- \( V_2 = ? \)

Solve for \( V_2 \):

\[
V_2 = \frac{P_1 V_1}{P_2} = \frac{(1 \text{ atm})(10 \text{ m}^3)}{0.3 \text{ atm}} = \frac{10}{0.3} = 33.\overline{3} \text{ m}^3
\]

So,
\[
V_2 \approx 33.3 \text{ m}^3
\]

✔ Answer:
The new volume of the weather balloon at 10 km altitude is approximately 33.3 m³.

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✔ Final Answers Summary:

#### Question 1:
- Liquid under pressure:
- Volume: doesn't change noticeably
- Density: doesn't change noticeably
- Gas under pressure:
- Volume: decreases
- Density: increases

#### Question 2:
- Sketch: Draw a larger balloon.
- Explanation: The balloon expands at high altitude because external atmospheric pressure decreases, allowing the gas inside to expand.

#### Question 3:
- Using Boyle’s Law:
\[
V_2 = \frac{P_1 V_1}{P_2} = \frac{1 \times 10}{0.3} = 33.3 \text{ m}^3
\]
- New volume = 33.3 m³

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Let me know if you'd like a visual sketch description or help drawing it!