Question

The temperature of a gas changes from 765 K to 315 K. What is the final volume of the container if the original volume was 986 mL?
Assume that the number of moles and the pressure remain constant.
V=[?]mL

Answer 1

To solve the problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and the number of moles remain constant. The law can be expressed mathematically as:

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

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

From the problem:
- \( V_1 = 986 \, \text{mL} \)
- \( T_1 = 765 \, \text{K} \)
- \( T_2 = 315 \, \text{K} \)

We need to find \( V_2 \).

Rearranging the equation gives us:

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

Plugging in the known values:

\[
V_2 = 986 \, \text{mL} \cdot \frac{315 \, \text{K}}{765 \, \text{K}}
\]

Calculating the intermediate value:

\[
V_2 = 986 \, \text{mL} \cdot \frac{315}{765} \approx 986 \, \text{mL} \cdot 0.4118 \approx 406.01 \, \text{mL}
\]

Thus, rounding to a reasonable number of significant figures, we find:

\[
V_2 \approx 406 \, \text{mL}
\]

So, the final volume of the container is approximately **406 mL**.