Questions: Question 37 A gas is at 35.0°C and 2.50 L. What is the temperature of the gas if the volume is increased to 5.00 L? (A) 65.0°C (B) 615°C (C) 17.5°C (D) 343°C (E) 1.16°C

Transcript text: Question 37 A gas is at $35.0^{\circ} \mathrm{C}$ and 2.50 L . What is the temperature of the gas if the volume is increased to 5.00 L ? (A) $65.0^{\circ} \mathrm{C}$ (B) $615^{\circ} \mathrm{C}$ (C) $17.5^{\circ} \mathrm{C}$ (D) $343^{\circ} \mathrm{C}$ (E) $1.16^{\circ} \mathrm{C}$

Solution

Solution Steps

Step 1: Identify the Given Information

We are given:

Initial temperature, $ T_1 = 35.0^{\circ} \mathrm{C} $
Initial volume, $ V_1 = 2.50 \, \mathrm{L} $
Final volume, $ V_2 = 5.00 \, \mathrm{L} $

Step 2: Convert Initial Temperature to Kelvin

To use the gas laws, we need to convert the temperature from Celsius to Kelvin: \[ T_1 = 35.0 + 273.15 = 308.15 \, \mathrm{K} \]

Step 3: Apply Charles's Law

Charles's Law states that for a given amount of gas at constant pressure, the volume is directly proportional to its temperature: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Step 4: Solve for the Final Temperature in Kelvin

Rearrange the equation to solve for $ T_2 $: \[ T_2 = \frac{V_2 \cdot T_1}{V_1} \] Substitute the known values: \[ T_2 = \frac{5.00 \, \mathrm{L} \cdot 308.15 \, \mathrm{K}}{2.50 \, \mathrm{L}} = 616.30 \, \mathrm{K} \]

Step 5: Convert Final Temperature to Celsius

Convert the final temperature back to Celsius: \[ T_2 = 616.30 - 273.15 = 343.15^{\circ} \mathrm{C} \]