To determine the volume that a given amount of chlorine gas will occupy at [tex]\(-20^{\circ} C\)[/tex] from its volume at [tex]\(200^{\circ} C\)[/tex], we can use Charles's Law. Charles's Law states that for a given mass of an ideal gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (Kelvin). The formula is:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
where:
- [tex]\( V_1 \)[/tex] and [tex]\( V_2 \)[/tex] are the initial and final volumes, respectively.
- [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] are the initial and final absolute temperatures, respectively.
Given data:
- Initial temperature, [tex]\( T_1 = 200^{\circ} C \)[/tex]
- Final temperature, [tex]\( T_2 = -20^{\circ} C \)[/tex]
- Initial volume, [tex]\( V_1 = 972.70 \text{ mL} \)[/tex]
First, we need to convert the temperatures from Celsius to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(^{\circ} C) + 273.15 \][/tex]
Calculating the initial and final temperatures in Kelvin:
- Initial temperature: [tex]\( T_1 = 200 + 273.15 = 473.15 \text{ K} \)[/tex]
- Final temperature: [tex]\( T_2 = -20 + 273.15 = 253.15 \text{ K} \)[/tex]
Now we can use Charles's Law to find the final volume [tex]\( V_2 \)[/tex]:
[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]
Substituting in the known values:
[tex]\[
V_2 = 972.70 \text{ mL} \times \frac{253.15 \text{ K}}{473.15 \text{ K}}
\][/tex]
After calculating, we get:
[tex]\[
V_2 \approx 520.42 \text{ mL}
\][/tex]
Therefore, the volume of the chlorine gas at [tex]\(-20^{\circ} C\)[/tex] will be approximately [tex]\(520.42 \text{ mL}\)[/tex].
