What Celsius temperature, [tex]T_2[/tex], is required to change the volume of the gas sample in Part A from [tex]T_1 = 17^\circ \text{C}[/tex] and [tex]V_1 = 1860 \, \text{L}[/tex] to a volume of [tex]3720 \, \text{L}[/tex]? Assume no change in pressure or the amount of gas.

A. 68.5 °C
B. 98.2 °C
C. 132.7 °C
D. 157.4 °C

To find the final temperature (T₂), we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and the amount of gas are held constant. Mathematically, V₁/T₁ = V₂/T₂. We rearrange the equation to solve for T₂, giving us T₂ = (V₂/V₁) * T₁. Substituting the given values, we get T₂ = (3720 L / 1860 L) * 17°C = 2 * 17°C = 34°C.

However, this is the temperature change, not the final temperature. Therefore, to find the final temperature, we add the initial temperature change to the initial temperature, giving us 17°C + 34°C = 51°C. However, this is in Celsius, and the question asks for Celsius, so the final answer is 132.7 °C (Option C). This temperature will double the volume of the gas sample while keeping the pressure and amount of gas constant, according to Charles's Law.