To solve this problem, we can use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature (in Kelvin), provided the volume remains constant:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where:

• $P_1$ and $T_1$ are the initial pressure and temperature.
• $P_2$ and $T_2$ are the final pressure and temperature.

Step 1: Convert temperatures to Kelvin.

The initial temperature is 27°C. To convert Celsius to Kelvin: $T_1 = 27 + 273.15 = 300.15 \text{ K}$

Step 2: Use the equation to find $T_2$.

Rearrange the equation to solve for $T_2$: $T_2 = \frac{P_2 \cdot T_1}{P_1}$

Substitute the known values:

• $P_1 = 450 \text{ mm Hg}$
• $P_2 = 200 \text{ mm Hg}$
• $T_1 = 300.15 \text{ K}$

$T_2 = \frac{200 \cdot 300.15}{450}$ $T_2 = 133.4 \text{ K}$

Step 3: Convert $T_2$ back to Celsius.

$T_2 = 133.4 - 273.15 = -139.75^\circ \text{C}$

Therefore, the temperature at which the gas's pressure will be 200 mm Hg is approximately -139.75°C.

Do you want more details or have any questions?

Related Questions:

1. How does Gay-Lussac's Law apply to real-life situations?
2. What would happen to the temperature if the pressure were increased instead of decreased?
3. Can Gay-Lussac's Law be used for gases that deviate from ideal behavior?
4. How would the calculation change if the volume were not held constant?
5. What is the relationship between pressure and temperature in gases according to the kinetic molecular theory?

Tip: Always remember to convert temperatures to Kelvin when using gas laws, as these laws require absolute temperature values.