Answer :
To determine the percentage increase in temperature necessary for a 10% increase in volume at constant pressure, we can use Charles's Law. Charles's Law states that for a given mass of an ideal gas at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin.
Mathematically, Charles's Law is 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,
- [tex]\( V_2 \)[/tex] is the final volume,
- [tex]\( T_2 \)[/tex] is the final temperature.
Given that the volume increases by 10%, we can express [tex]\( V_2 \)[/tex] as:
[tex]\[ V_2 = 1.10 \times V_1 \][/tex]
We need to find the percentage increase in temperature, which means finding [tex]\( T_2 \)[/tex].
First, rearrange Charles's Law to solve for [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = T_1 \times \frac{V_2}{V_1} \][/tex]
Substitute [tex]\( V_2 = 1.10 \times V_1 \)[/tex] into the equation:
[tex]\[ T_2 = T_1 \times \frac{1.10 \times V_1}{V_1} \][/tex]
[tex]\[ T_2 = T_1 \times 1.10 \][/tex]
Now, the percentage increase in temperature can be calculated as:
[tex]\[ \text{Percentage increase in temperature} = \left(\frac{T_2 - T_1}{T_1}\right) \times 100\% \][/tex]
Substitute [tex]\( T_2 = 1.10 \times T_1 \)[/tex] into the equation:
[tex]\[ \text{Percentage increase in temperature} = \left(\frac{1.10 \times T_1 - T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = \left(\frac{1.10T_1 - T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = \left(\frac{0.10T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = 0.10 \times 100\% \][/tex]
[tex]\[ = 10\% \][/tex]
Therefore, the percentage increase in temperature required for a 10% increase in volume at constant pressure is:
10%
So, the correct answer is C (10%).
Mathematically, Charles's Law is 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,
- [tex]\( V_2 \)[/tex] is the final volume,
- [tex]\( T_2 \)[/tex] is the final temperature.
Given that the volume increases by 10%, we can express [tex]\( V_2 \)[/tex] as:
[tex]\[ V_2 = 1.10 \times V_1 \][/tex]
We need to find the percentage increase in temperature, which means finding [tex]\( T_2 \)[/tex].
First, rearrange Charles's Law to solve for [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = T_1 \times \frac{V_2}{V_1} \][/tex]
Substitute [tex]\( V_2 = 1.10 \times V_1 \)[/tex] into the equation:
[tex]\[ T_2 = T_1 \times \frac{1.10 \times V_1}{V_1} \][/tex]
[tex]\[ T_2 = T_1 \times 1.10 \][/tex]
Now, the percentage increase in temperature can be calculated as:
[tex]\[ \text{Percentage increase in temperature} = \left(\frac{T_2 - T_1}{T_1}\right) \times 100\% \][/tex]
Substitute [tex]\( T_2 = 1.10 \times T_1 \)[/tex] into the equation:
[tex]\[ \text{Percentage increase in temperature} = \left(\frac{1.10 \times T_1 - T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = \left(\frac{1.10T_1 - T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = \left(\frac{0.10T_1}{T_1}\right) \times 100\% \][/tex]
[tex]\[ = 0.10 \times 100\% \][/tex]
[tex]\[ = 10\% \][/tex]
Therefore, the percentage increase in temperature required for a 10% increase in volume at constant pressure is:
10%
So, the correct answer is C (10%).
