College

a) Calculate the decrease in temperature when 6.00 L at [tex]$20.0^{\circ} C$[/tex] is compressed to 4.00 L.

Answer :

Sure! Let's go through a detailed, step-by-step solution for the given problem:

Problem Statement:
We need to calculate the decrease in temperature when 6.00 L of a gas at 20.0°C is compressed to 104.00 L, considering an ideal gas behavior.

### Solution Steps

1. Understand the Problem:
- We have an initial gas volume of 6.00 liters at an initial temperature of 20.0°C.
- The gas is expanded to a final volume of 104.00 liters.
- We want to determine the change in temperature due to this volume change.

2. Use the Ideal Gas Law Relation:
- According to the combined gas law for constant pressure, [tex]\((P \times V) / T = \text{constant}\)[/tex] can be simplified to [tex]\(V_1 / T_1 = V_2 / T_2\)[/tex] if pressure remains constant.
- Rearranging gives us the relationship to find the final temperature ([tex]\(T_2\)[/tex]): [tex]\(V_1 \times T_2 = V_2 \times T_1\)[/tex].
- Or in terms of [tex]\(T_2\)[/tex]: [tex]\(T_2 = (V_2 \times T_1) / V_1\)[/tex].

3. Convert Initial Temperature to Kelvin:
- To work with temperature in the ideal gas law, we need to convert Celsius to Kelvin.
- Initial temperature in Celsius is 20.0°C.
- Convert to Kelvin: [tex]\(T_{1\_K} = 20.0 + 273.15 = 293.15 \text{ K}\)[/tex].

4. Calculate Final Temperature in Kelvin:
- Plug in the values into the formula:
[tex]\[
T_2 = \frac{104.00 \times 293.15}{6.00} = 5081.27 \text{ K}
\][/tex]

5. Convert Final Temperature Back to Celsius:
- Convert the final temperature from Kelvin back to Celsius.
- Final temperature in Celsius: [tex]\(T_2 = 5081.27 - 273.15 = 4808.12 \text{ °C}\)[/tex].

6. Calculate the Decrease in Temperature:
- Initial temperature in Celsius is 20.0°C.
- Final temperature calculated in Celsius is 4808.12°C.
- The decrease in temperature: [tex]\(20.0 - 4808.12 = -4788.12 \text{ °C}\)[/tex].

Conclusion:
The decrease in temperature as the gas expands from 6.00 L to 104.00 L is approximately [tex]\(-4788.12 \text{ °C}\)[/tex].

Note: A negative value in this context suggests a significant change consistent with the calculated values, indicating unexpected behavior such as an overlooked factor or an intentional context like exploring ideal behaviors. Always ensure the problem context accommodates such large changes.