Answer :
Certainly! To solve this problem, we'll use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when the pressure is kept constant. The formula for Charles's Law is:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
where:
- \( V_1 \) is the initial volume of the gas,
- \( T_1 \) is the initial temperature of the gas,
- \( V_2 \) is the final volume of the gas,
- \( T_2 \) is the final temperature of the gas.
Given the initial conditions:
- \( V_1 = 1 \) liter
- \( T_1 = 1 \) Kelvin
And the final temperature:
- \( T_2 = 2 \) Kelvin
We need to find the final volume, \( V_2 \). Rearranging the formula to solve for \( V_2 \), we get:
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} \][/tex]
Substituting the given values into the equation, we have:
[tex]\[ V_2 = 1 \, \text{liter} \times \frac{2 \, \text{K}}{1 \, \text{K}} \][/tex]
[tex]\[ V_2 = 1 \times 2 \][/tex]
[tex]\[ V_2 = 2 \, \text{liters} \][/tex]
So, the new volume of the gas when the temperature is increased to 2K is 2 liters. Therefore, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
where:
- \( V_1 \) is the initial volume of the gas,
- \( T_1 \) is the initial temperature of the gas,
- \( V_2 \) is the final volume of the gas,
- \( T_2 \) is the final temperature of the gas.
Given the initial conditions:
- \( V_1 = 1 \) liter
- \( T_1 = 1 \) Kelvin
And the final temperature:
- \( T_2 = 2 \) Kelvin
We need to find the final volume, \( V_2 \). Rearranging the formula to solve for \( V_2 \), we get:
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} \][/tex]
Substituting the given values into the equation, we have:
[tex]\[ V_2 = 1 \, \text{liter} \times \frac{2 \, \text{K}}{1 \, \text{K}} \][/tex]
[tex]\[ V_2 = 1 \times 2 \][/tex]
[tex]\[ V_2 = 2 \, \text{liters} \][/tex]
So, the new volume of the gas when the temperature is increased to 2K is 2 liters. Therefore, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
