Question

a sample of gas originally occupies 17.5 l at -23.3 °c. what is its new volume, in l, when it is heated to 38.5 °c? report your answer to the correct number of significant figures.

Explanation:

Step1: Convert temperatures to Kelvin

To use Charles's Law, we first convert the Celsius temperatures to Kelvin. The formula to convert Celsius to Kelvin is \( T(K) = T(^\circ C) + 273.15 \).

For the initial temperature \( T_1 \):
\( T_1 = -23.3^\circ C + 273.15 = 249.85 \, K \)

For the final temperature \( T_2 \):
\( T_2 = 38.5^\circ C + 273.15 = 311.65 \, K \)

Step2: Apply Charles's Law

Charles's Law states that for a fixed amount of gas at constant pressure, \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V_1 \) is the initial volume, \( T_1 \) is the initial temperature, \( V_2 \) is the final volume, and \( T_2 \) is the final temperature.

We know \( V_1 = 17.5 \, L \), \( T_1 = 249.85 \, K \), and \( T_2 = 311.65 \, K \). We need to solve for \( V_2 \).

Rearranging the formula for \( V_2 \):
\( V_2 = V_1 \times \frac{T_2}{T_1} \)

Substitute the known values:
\( V_2 = 17.5 \, L \times \frac{311.65 \, K}{249.85 \, K} \)

Step3: Calculate the final volume

First, calculate the ratio \( \frac{311.65}{249.85} \approx 1.247 \)

Then, multiply by \( V_1 \):
\( V_2 = 17.5 \, L \times 1.247 \approx 21.82 \, L \)

Considering significant figures, the initial volume (17.5 L) has three significant figures, and the temperatures ( -23.3 and 38.5) have three significant figures, so the result should have three significant figures.

Answer:

\( \approx 21.8 \, L \) (or more precisely \( \approx 21.8 \) when rounded to three significant figures)