QUESTION 17

If 3.0 L of helium at 20°C is allowed to expand to 4.4 L, with the pressure remaining the same, what is the new temperature?

A. 102°C
B. 150°C
C. 157°C
D. 230°C



Answer :

To solve this problem, we need to use the principles of the ideal gas law, specifically the relationship between volume and temperature at constant pressure. This relationship can be expressed as:

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

Where:
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin
- [tex]\( V_2 \)[/tex] is the final volume
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin

Given:
- Initial volume ([tex]\( V_1 \)[/tex]) = 3.0 liters
- Initial temperature ([tex]\( T_1 \)[/tex]) = 20°C

First, we need to convert the initial temperature from Celsius to Kelvin, since the ideal gas law requires absolute temperatures:
[tex]\[ T_1 = 20 + 273.15 = 293.15 \text{ K} \][/tex]

The final volume ([tex]\( V_2 \)[/tex]) is given as 4.4 liters. We need to find the final temperature ([tex]\( T_2 \)[/tex]).

Using the formula:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]

Plugging in the values:
[tex]\[ \frac{3.0 \, \text{L}}{293.15 \, \text{K}} = \frac{4.4 \, \text{L}}{T_2} \][/tex]

Solving for [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = \frac{4.4 \times 293.15}{3.0} \][/tex]
[tex]\[ T_2 \approx 429.953 \, \text{K} \][/tex]

Finally, convert the final temperature from Kelvin back to Celsius:
[tex]\[ T_2 = 429.953 \, \text{K} - 273.15 \][/tex]
[tex]\[ T_2 \approx 156.803 \, \text{°C} \][/tex]

Therefore, the new temperature is approximately 157°C. The correct answer is:
C. 157°C