Charles' Law Worksheet featuring four problems on gas behavior under changing temperatures.

It looks like you've uploaded a worksheet titled "Charles' Law Worksheet" with four problems related to Charles' Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure and amount of gas are held constant. The formula for Charles' Law is:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]

Where:
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature in Kelvin
- \( V_2 \) = final volume
- \( T_2 \) = final temperature in Kelvin

Let’s solve each problem step by step.

---

Problem 1:

> The temperature inside my balloon is about 27°C. If I place a balloon in my fridge that initially has a temperature of 27°C and a volume of 5.0 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?

We need the temperature of the refrigerator. Since it's not given, we assume a typical refrigerator temperature of 4°C (which is common). But since this isn't stated, let's suppose it's 4°C unless otherwise specified.

But wait — the problem says "fully cooled by my refrigerator" but doesn’t give a specific temperature. This might be an error or missing information.

Wait — actually, looking back: It says "initially has a temperature of 27°C" and then is placed in the fridge. But no final temperature is given. So perhaps this is incomplete? Or maybe it's implied?

Alternatively, maybe there's a typo and the second part is meant to be solved with a different temperature.

Wait — let me re-read carefully:

> "The temperature inside my balloon is about 27°C. If I place a balloon in my fridge that initially has a temperature of 27°C and a volume of 5.0 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?"

So the balloon starts at 27°C and 5.0 L. Then it's cooled in the fridge. But we don’t know the fridge temperature.

This is a missing value. Unless the fridge temperature is standard, we can’t proceed.

But maybe the user intended to provide the fridge temperature, or perhaps it's assumed to be 0°C or 4°C.

Let’s assume fridge temperature = 4°C (a typical refrigerator temperature).

Convert temperatures to Kelvin:

- \( T_1 = 27^\circ C = 27 + 273 = 300\,K \)
- \( T_2 = 4^\circ C = 4 + 273 = 277\,K \)
- \( V_1 = 5.0\,L \)

Using Charles’ Law:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow V_2 = V_1 \times \frac{T_2}{T_1}
\]

\[
V_2 = 5.0 \times \frac{277}{300} = 5.0 \times 0.9233 = 4.6165\,L
\]

\[
\boxed{V_2 \approx 4.62\,L}
\]

✔ Answer: ~4.62 L

> ⚠️ Note: This assumes fridge temperature is 4°C. If a different temperature was intended, please specify.

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Problem 2:

> A man heats a balloon in the oven. If the balloon initially has a volume of 5.4 liters and a temperature of 20°C, what will the volume of the balloon be after he heats it to a temperature of 100°C?

Convert temperatures to Kelvin:

- \( T_1 = 20^\circ C = 293\,K \)
- \( T_2 = 100^\circ C = 373\,K \)
- \( V_1 = 5.4\,L \)

\[
V_2 = V_1 \times \frac{T_2}{T_1} = 5.4 \times \frac{373}{293}
\]

Calculate:

\[
\frac{373}{293} \approx 1.273
\]
\[
V_2 = 5.4 \times 1.273 \approx 6.874\,L
\]

\[
\boxed{V_2 \approx 6.87\,L}
\]

✔ Answer: ~6.87 L

---

Problem 3:

> On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I leave a 250 mL bag at a temperature of 19°C, and I leave it in my car which has a temperature of 65°C, what will the new volume of the bag be?

Use Charles’ Law again.

- \( V_1 = 250\,mL \)
- \( T_1 = 19^\circ C = 19 + 273 = 292\,K \)
- \( T_2 = 65^\circ C = 65 + 273 = 338\,K \)

\[
V_2 = 250 \times \frac{338}{292}
\]

\[
\frac{338}{292} \approx 1.1575
\]
\[
V_2 = 250 \times 1.1575 = 289.375\,mL
\]

\[
\boxed{V_2 \approx 289\,mL}
\]

✔ Answer: ~289 mL

---

Problem 4:

> A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an empty soda bottle (of 2.1 L) at room temperature (25°C), what will the new volume be if you put it in your freezer (-10°C)?

Convert temperatures:

- \( T_1 = 25^\circ C = 298\,K \)
- \( T_2 = -10^\circ C = 263\,K \)
- \( V_1 = 2.1\,L \)

\[
V_2 = 2.1 \times \frac{263}{298}
\]

\[
\frac{263}{298} \approx 0.8825
\]
\[
V_2 = 2.1 \times 0.8825 = 1.85325\,L
\]

\[
\boxed{V_2 \approx 1.85\,L}
\]

✔ Answer: ~1.85 L

---

✔ Final Answers Summary:

1. ~4.62 L (assuming fridge temp = 4°C)
2. ~6.87 L
3. ~289 mL
4. ~1.85 L

> 🔍 Note: Problem 1 requires an assumption about fridge temperature. If the fridge temperature was meant to be something else (e.g., 0°C), the answer would differ slightly. For example, if fridge = 0°C = 273 K:

\[
V_2 = 5.0 \times \frac{273}{300} = 5.0 \times 0.91 = 4.55\,L
\]

So always check the given values!

Let me know if you'd like these answers formatted neatly or printed out!