To determine the new volume of a gas when it is cooled from [tex]\(37.0^\circ \text{C}\)[/tex] to [tex]\(0.0^\circ \text{C}\)[/tex], we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. Mathematically, Charles's Law can be expressed as:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Here:
- [tex]\(V_1\)[/tex] is the initial volume of the gas.
- [tex]\(T_1\)[/tex] is the initial temperature of the gas in Kelvin.
- [tex]\(V_2\)[/tex] is the final volume of the gas.
- [tex]\(T_2\)[/tex] is the final temperature of the gas in Kelvin.
Given:
- Initial volume, [tex]\(V_1 = 50.0 \text{ mL}\)[/tex]
- Initial temperature, [tex]\(T_1 = 37.0^\circ \text{C}\)[/tex]
- Final temperature, [tex]\(T_2 = 0.0^\circ \text{C}\)[/tex]
First, we need to convert the temperatures from Celsius to Kelvin. The conversion formula is:
[tex]\[
T \text{ (K)} = T \text{ (°C)} + 273.15
\][/tex]
So:
[tex]\[
T_1 = 37.0 + 273.15 = 310.15 \text{ K}
\][/tex]
[tex]\[
T_2 = 0.0 + 273.15 = 273.15 \text{ K}
\][/tex]
Now, using Charles's Law:
[tex]\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\][/tex]
Rearranging to solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = V_1 \times \frac{T_2}{T_1}
\][/tex]
Substitute in the known values:
[tex]\[
V_2 = 50.0 \text{ mL} \times \frac{273.15 \text{ K}}{310.15 \text{ K}}
\][/tex]
Multiplying through:
[tex]\[
V_2 \approx 50.0 \text{ mL} \times 0.880702
\][/tex]
[tex]\[
V_2 \approx 44.04 \text{ mL}
\][/tex]
Therefore, when a 50.0 mL sample of gas at [tex]\(37.0^\circ \text{C}\)[/tex] is cooled to [tex]\(0.0^\circ \text{C}\)[/tex], the new volume of the gas is approximately 44.04 mL.
