If a sample of gas at constant pressure has a volume of 441 mL at 36.5 °C, what will its volume be if the temperature is increased to 77.6 °C ?

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**Gas Behavior Under Temperature Change**

**Problem Statement:**

If a sample of gas at constant pressure has a volume of 441 mL at 36.5 °C, what will its volume be if the temperature is increased to 77.6 °C?

This question relates to the principles of gas behavior under constant pressure, typically described by Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature when measured in Kelvin. To solve this problem, we'll use the following formula derived from Charles's Law:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where:
- \( V_1 \) is the initial volume (441 mL),
- \( T_1 \) is the initial temperature in Kelvin (36.5 °C converted to Kelvin),
- \( V_2 \) is the final volume (unknown),
- \( T_2 \) is the final temperature in Kelvin (77.6 °C converted to Kelvin).

**Step-by-Step Solution:**

1. Convert the temperatures from Celsius to Kelvin:
\[ T_1 = 36.5 + 273.15 = 309.65 \text{ K} \]
\[ T_2 = 77.6 + 273.15 = 350.75 \text{ K} \]

2. Apply Charles's Law:
\[ \frac{441 \text{ mL}}{309.65 \text{ K}} = \frac{V_2}{350.75 \text{ K}} \]

3. Solve for \( V_2 \):
\[ V_2 = \frac{441 \text{ mL} \times 350.75 \text{ K}}{309.65 \text{ K}} \]
\[ V_2 \approx 499.15 \text{ mL} \]

Thus, if the temperature of the gas is increased to 77.6 °C, the volume of the gas will increase to approximately 499.15 mL.
Transcribed Image Text:**Gas Behavior Under Temperature Change** **Problem Statement:** If a sample of gas at constant pressure has a volume of 441 mL at 36.5 °C, what will its volume be if the temperature is increased to 77.6 °C? This question relates to the principles of gas behavior under constant pressure, typically described by Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature when measured in Kelvin. To solve this problem, we'll use the following formula derived from Charles's Law: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where: - \( V_1 \) is the initial volume (441 mL), - \( T_1 \) is the initial temperature in Kelvin (36.5 °C converted to Kelvin), - \( V_2 \) is the final volume (unknown), - \( T_2 \) is the final temperature in Kelvin (77.6 °C converted to Kelvin). **Step-by-Step Solution:** 1. Convert the temperatures from Celsius to Kelvin: \[ T_1 = 36.5 + 273.15 = 309.65 \text{ K} \] \[ T_2 = 77.6 + 273.15 = 350.75 \text{ K} \] 2. Apply Charles's Law: \[ \frac{441 \text{ mL}}{309.65 \text{ K}} = \frac{V_2}{350.75 \text{ K}} \] 3. Solve for \( V_2 \): \[ V_2 = \frac{441 \text{ mL} \times 350.75 \text{ K}}{309.65 \text{ K}} \] \[ V_2 \approx 499.15 \text{ mL} \] Thus, if the temperature of the gas is increased to 77.6 °C, the volume of the gas will increase to approximately 499.15 mL.
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