Answer :
To solve the problem of finding the final volume of ethane gas when it's heated from 25.0 °C to 50.0 °C at constant pressure, we can use Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature in Kelvin, as long as the pressure remains constant. The formula is:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the initial volume (40.0 mL).
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin.
- [tex]\( V_2 \)[/tex] is the final volume.
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin.
### Step-by-Step Solution:
1. Convert Temperatures to Kelvin:
- Initial temperature ([tex]\( T_1 \)[/tex]) = 25.0 °C. To convert to Kelvin, use:
[tex]\[
T_1 = 25.0 + 273.15 = 298.15 \, \text{K}
\][/tex]
- Final temperature ([tex]\( T_2 \)[/tex]) = 50.0 °C. To convert to Kelvin, use:
[tex]\[
T_2 = 50.0 + 273.15 = 323.15 \, \text{K}
\][/tex]
2. Apply Charles's Law:
Substitute the known values into the formula:
[tex]\[
\frac{40.0 \, \text{mL}}{298.15 \, \text{K}} = \frac{V_2}{323.15 \, \text{K}}
\][/tex]
3. Solve for [tex]\( V_2 \)[/tex]:
Rearrange the equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = 40.0 \, \text{mL} \times \frac{323.15 \, \text{K}}{298.15 \, \text{K}}
\][/tex]
Doing the calculation gives:
[tex]\[
V_2 \approx 43.4 \, \text{mL}
\][/tex]
Therefore, the final volume of the ethane gas when heated to 50.0 °C at constant pressure is approximately 43.4 mL. So, the answer is 43.4 mL.
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the initial volume (40.0 mL).
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin.
- [tex]\( V_2 \)[/tex] is the final volume.
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin.
### Step-by-Step Solution:
1. Convert Temperatures to Kelvin:
- Initial temperature ([tex]\( T_1 \)[/tex]) = 25.0 °C. To convert to Kelvin, use:
[tex]\[
T_1 = 25.0 + 273.15 = 298.15 \, \text{K}
\][/tex]
- Final temperature ([tex]\( T_2 \)[/tex]) = 50.0 °C. To convert to Kelvin, use:
[tex]\[
T_2 = 50.0 + 273.15 = 323.15 \, \text{K}
\][/tex]
2. Apply Charles's Law:
Substitute the known values into the formula:
[tex]\[
\frac{40.0 \, \text{mL}}{298.15 \, \text{K}} = \frac{V_2}{323.15 \, \text{K}}
\][/tex]
3. Solve for [tex]\( V_2 \)[/tex]:
Rearrange the equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = 40.0 \, \text{mL} \times \frac{323.15 \, \text{K}}{298.15 \, \text{K}}
\][/tex]
Doing the calculation gives:
[tex]\[
V_2 \approx 43.4 \, \text{mL}
\][/tex]
Therefore, the final volume of the ethane gas when heated to 50.0 °C at constant pressure is approximately 43.4 mL. So, the answer is 43.4 mL.
