Math Snap

STEP 1

What is this asking?
If a balloon starts at $5.0$ L at $20.0^\circ \text{C}$, how big will it get at $-196^\circ \text{C}$?
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 $273.15$ to our Celsius temperature.
So, for our initial temperature:
$$20.0^\circ \text{C} + 273.15 = \textbf{293.15 K}$$

STEP 5

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

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:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$Where $V_1$ and $T_1$ are the initial volume and initial temperature, and $V_2$ and $T_2$ are the final volume and final temperature.

STEP 7

Let's plug in our values!
Our initial volume $V_1$ is $5.0$ L, and our initial temperature $T_1$ is $\textbf{293.15 K}$.
Our final temperature $T_2$ is $\textbf{77.15 K}$, and we're trying to find $V_2$, the final volume.
$$\frac{5.0 \text{ L}}{293.15 \text{ K}} = \frac{V_2}{77.15 \text{ K}}$$

STEP 8

To solve for $V_2$, we can multiply both sides of the equation by $77.15 \text{ K}$:
$$\frac{5.0 \text{ L}}{293.15 \text{ K}} \cdot 77.15 \text{ K} = V_2$$

STEP 9

Now, we just crunch the numbers:
$$V_2 = \frac{5.0 \text{ L} \cdot 77.15}{293.15} = \frac{385.75}{293.15} \text{ L} \approx \textbf{1.3 L}$$

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