Answer :
To solve this problem, we will use the principles of the Ideal Gas Law, specifically Charles' Law, which states that the volume of a gas is directly proportional to its temperature when the pressure and the number of moles remain constant. The formula can be expressed as:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Here, [tex]\(V_1\)[/tex] and [tex]\(T_1\)[/tex] are the initial volume and temperature, and [tex]\(V_2\)[/tex] and [tex]\(T_2\)[/tex] are the final volume and temperature, respectively.
1. Given data:
- Initial volume, [tex]\(V_1 = 800.0\)[/tex] mL
- Initial temperature, [tex]\(T_1 = 115^{\circ} C\)[/tex]
- Final volume, [tex]\(V_2 = 400.0\)[/tex] mL
2. Convert the initial temperature to Kelvin:
To use the gas law properly, we need to work in absolute temperatures (Kelvin). We convert the initial temperature from Celsius to Kelvin using the conversion formula:
[tex]\[ T (\text{K}) = T (\text{C}) + 273.15 \][/tex]
So,
[tex]\[ T_1 = 115 + 273.15 = 388.15 \text{ K} \][/tex]
3. Apply Charles' Law:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Rearranging the formula to find the final temperature [tex]\(T_2\)[/tex] we get:
[tex]\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \][/tex]
Substituting the given values:
[tex]\[ T_2 = \frac{400.0 \text{ mL} \cdot 388.15 \text{ K}}{800.0 \text{ mL}} \][/tex]
[tex]\[ T_2 = 194.075 \text{ K} \][/tex]
4. Convert [tex]\(T_2\)[/tex] back to Celsius:
To find the temperature in Celsius, we convert back using:
[tex]\[ T (\text{C}) = T (\text{K}) - 273.15 \][/tex]
Therefore,
[tex]\[ T_2 = 194.075 - 273.15 = -79.075^{\circ} C \][/tex]
The new temperature when the gas volume shrinks to 400.0 mL is [tex]\(-79.08^{\circ} C\)[/tex] (rounded to two decimal places).
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Here, [tex]\(V_1\)[/tex] and [tex]\(T_1\)[/tex] are the initial volume and temperature, and [tex]\(V_2\)[/tex] and [tex]\(T_2\)[/tex] are the final volume and temperature, respectively.
1. Given data:
- Initial volume, [tex]\(V_1 = 800.0\)[/tex] mL
- Initial temperature, [tex]\(T_1 = 115^{\circ} C\)[/tex]
- Final volume, [tex]\(V_2 = 400.0\)[/tex] mL
2. Convert the initial temperature to Kelvin:
To use the gas law properly, we need to work in absolute temperatures (Kelvin). We convert the initial temperature from Celsius to Kelvin using the conversion formula:
[tex]\[ T (\text{K}) = T (\text{C}) + 273.15 \][/tex]
So,
[tex]\[ T_1 = 115 + 273.15 = 388.15 \text{ K} \][/tex]
3. Apply Charles' Law:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Rearranging the formula to find the final temperature [tex]\(T_2\)[/tex] we get:
[tex]\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \][/tex]
Substituting the given values:
[tex]\[ T_2 = \frac{400.0 \text{ mL} \cdot 388.15 \text{ K}}{800.0 \text{ mL}} \][/tex]
[tex]\[ T_2 = 194.075 \text{ K} \][/tex]
4. Convert [tex]\(T_2\)[/tex] back to Celsius:
To find the temperature in Celsius, we convert back using:
[tex]\[ T (\text{C}) = T (\text{K}) - 273.15 \][/tex]
Therefore,
[tex]\[ T_2 = 194.075 - 273.15 = -79.075^{\circ} C \][/tex]
The new temperature when the gas volume shrinks to 400.0 mL is [tex]\(-79.08^{\circ} C\)[/tex] (rounded to two decimal places).
