Question

At atmospheric pressure, the temperature of a gas that occupies a volume of 5.30L is 298K. What is the temperature of a gas if it is allowed to expand to 7.60L

Answer 1

To determine the new temperature of the gas after it expands at constant atmospheric pressure, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is constant. The formula is given by:

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

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

Given:
- \( V_1 = 5.30 \, \text{L} \)
- \( T_1 = 298 \, \text{K} \)
- \( V_2 = 7.60 \, \text{L} \)

We need to find \( T_2 \).

Rearranging Charles's Law gives us:

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

Now, substituting in the known values:

\[
T_2 = 298 \, \text{K} \times \frac{7.60 \, \text{L}}{5.30 \, \text{L}}
\]

Calculating the fraction:

\[
T_2 = 298 \, \text{K} \times \frac{7.60}{5.30} \approx 298 \, \text{K} \times 1.43 \approx 426.74 \, \text{K}
\]

So, the final temperature \( T_2 \) is approximately **426.74 K**.