Math  /  Algebra

QuestionIf the balloon in the video had an initial volume of 5.0 L at room temperature (20.0C)\left(20.0^{\circ} \mathrm{C}\right), then what would be the volume of the balloon after it was placed in liquid nitrogen with a temperature of 196C-196^{\circ} \mathrm{C} ? a. 49 L b. 2.0 L c. 4.6 L d. 1.3 L e. 19 L

Studdy Solution

STEP 1

What is this asking? If a balloon starts at 5.05.0 L at 20.0C20.0^\circ \text{C}, how big will it get at 196C-196^\circ \text{C}? Watch out! Don't forget to convert Celsius to Kelvin!

STEP 2

1. Convert temperatures to Kelvin.
2. Calculate the new volume.

STEP 3

Alright students, let's **kick things off** by converting those Celsius temperatures to Kelvin!
Remember, we *always* use Kelvin for gas law problems!
Why? Because Kelvin starts at **absolute zero**, the coldest possible temperature, which makes our calculations much more accurate and predictable.

STEP 4

To convert Celsius to Kelvin, we just **add** 273.15273.15 to our Celsius temperature.
So, for our **initial temperature**:
20.0C+273.15=293.15 K20.0^\circ \text{C} + 273.15 = \textbf{293.15 K}

STEP 5

And for the **liquid nitrogen temperature**:
196C+273.15=77.15 K-196^\circ \text{C} + 273.15 = \textbf{77.15 K}

STEP 6

Now, we'll use Charles's Law, which says that the volume of a gas is directly proportional to its temperature when the pressure is constant.
This means if the temperature goes down, the volume goes down too!
The formula is:
V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}Where V1V_1 and T1T_1 are the **initial volume** and **initial temperature**, and V2V_2 and T2T_2 are the **final volume** and **final temperature**.

STEP 7

Let's **plug in** our values!
Our initial volume V1V_1 is 5.05.0 L, and our initial temperature T1T_1 is 293.15 K\textbf{293.15 K}.
Our final temperature T2T_2 is 77.15 K\textbf{77.15 K}, and we're trying to find V2V_2, the final volume.
5.0 L293.15 K=V277.15 K\frac{5.0 \text{ L}}{293.15 \text{ K}} = \frac{V_2}{77.15 \text{ K}}

STEP 8

To **solve for** V2V_2, we can **multiply both sides** of the equation by 77.15 K77.15 \text{ K}:
5.0 L293.15 K77.15 K=V2\frac{5.0 \text{ L}}{293.15 \text{ K}} \cdot 77.15 \text{ K} = V_2

STEP 9

Now, we just **crunch the numbers**:
V2=5.0 L77.15293.15=385.75293.15 L1.3 LV_2 = \frac{5.0 \text{ L} \cdot 77.15}{293.15} = \frac{385.75}{293.15} \text{ L} \approx \textbf{1.3 L}

STEP 10

So, the volume of the balloon after being placed in liquid nitrogen would be approximately 1.3 L\textbf{1.3 L}, which matches answer choice *d*.

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