Answer :
To solve the problem, we will use Charles's law, which states that for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature. This can be written as:
[tex]\[ \frac{T_1}{V_1} = \frac{T_2}{V_2} \][/tex]
Where:
- \( T_1 \) is the initial temperature,
- \( V_1 \) is the initial volume,
- \( T_2 \) is the final temperature,
- \( V_2 \) is the final volume.
Given:
- Initial temperature, \( T_1 = 25 \) degrees Celsius,
- Initial volume, \( V_1 = 250 \) cubic meters,
- Final volume, \( V_2 = 300 \) cubic meters.
First, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(C) + 273.15 \][/tex]
Thus,
[tex]\[ T_1(K) = 25 + 273.15 = 298.15 \text{ K} \][/tex]
Next, we can use Charles’s law to find the final temperature \( T_2 \) in Kelvin:
[tex]\[ \frac{T_1}{V_1} = \frac{T_2}{V_2} \][/tex]
Rearranging to solve for \( T_2 \):
[tex]\[ T_2 = \frac{T_1 \cdot V_2}{V_1} \][/tex]
Substituting the known values:
[tex]\[ T_2 = \frac{298.15 \text{ K} \cdot 300 \text{ m}^3}{250 \text{ m}^3} = 357.78 \text{ K} \][/tex]
Finally, convert the final temperature back to degrees Celsius:
[tex]\[ T_2(C) = T_2(K) - 273.15 \][/tex]
[tex]\[ T_2(C) = 357.78 - 273.15 = 84.63 \text{ degrees Celsius} \][/tex]
So, the new temperature when the volume is 300 cubic meters is approximately 84.63 degrees Celsius.
[tex]\[ \frac{T_1}{V_1} = \frac{T_2}{V_2} \][/tex]
Where:
- \( T_1 \) is the initial temperature,
- \( V_1 \) is the initial volume,
- \( T_2 \) is the final temperature,
- \( V_2 \) is the final volume.
Given:
- Initial temperature, \( T_1 = 25 \) degrees Celsius,
- Initial volume, \( V_1 = 250 \) cubic meters,
- Final volume, \( V_2 = 300 \) cubic meters.
First, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(C) + 273.15 \][/tex]
Thus,
[tex]\[ T_1(K) = 25 + 273.15 = 298.15 \text{ K} \][/tex]
Next, we can use Charles’s law to find the final temperature \( T_2 \) in Kelvin:
[tex]\[ \frac{T_1}{V_1} = \frac{T_2}{V_2} \][/tex]
Rearranging to solve for \( T_2 \):
[tex]\[ T_2 = \frac{T_1 \cdot V_2}{V_1} \][/tex]
Substituting the known values:
[tex]\[ T_2 = \frac{298.15 \text{ K} \cdot 300 \text{ m}^3}{250 \text{ m}^3} = 357.78 \text{ K} \][/tex]
Finally, convert the final temperature back to degrees Celsius:
[tex]\[ T_2(C) = T_2(K) - 273.15 \][/tex]
[tex]\[ T_2(C) = 357.78 - 273.15 = 84.63 \text{ degrees Celsius} \][/tex]
So, the new temperature when the volume is 300 cubic meters is approximately 84.63 degrees Celsius.
