Import Question JSON

$\text{According to the given problem, the gas obeys the following law:}$ \\ $\text{VP}^2 = \text{constant}$ \\ $\text{This means that for two different states (initial and final), we have:}$ \\ $\text{V}_1\text{P}_1^2 = \text{V}_2\text{P}_2^2$ \\ $\text{The ideal gas law is given by:}$ \\ $\text{PV} = \text{nRT}$ \\ $\text{From this equation, we can express pressure } \text{P} \text{ in terms of volume } \text{V} \text{ and temperature } \text{T}:$ \\ $\text{P} = \frac{\text{nRT}}{\text{V}}$ \\ $\text{Substitute the expression for } \text{P} \text{ into the given law:}$ \\ $\text{V}\left(\frac{\text{nRT}}{\text{V}}\right)^2 = \text{constant}$ \\ $\text{V}\frac{\text{n}^2\text{R}^2\text{T}^2}{\text{V}^2} = \text{constant}$ \\ $\frac{\text{n}^2\text{R}^2\text{T}^2}{\text{V}} = \text{constant}$ \\ $\text{Since } \text{n} \text{ and } \text{R} \text{ are constants, the equation simplifies to:}$ \\ $\frac{\text{T}^2}{\text{V}} = \text{constant}$ \\ $\text{This means that for the initial and final states:}$ \\ $\frac{\text{T}_1^2}{\text{V}_1} = \frac{\text{T}_2^2}{\text{V}_2}$ \\ $\text{Given initial conditions: } \text{T}_1 = \text{T}, \text{V}_1 = \text{V}$ \\ $\text{Given final conditions: } \text{T}_2 = \text{T}', \text{V}_2 = 2\text{V}$ \\ $\text{Substitute these values into the derived equation:}$ \\ $\frac{\text{T}^2}{\text{V}} = \frac{(\text{T}')^2}{2\text{V}}$ \\ $\text{Cancel } \text{V} \text{ from both sides:}$ \\ $\text{T}^2 = \frac{(\text{T}')^2}{2}$ \\ $\text{Rearrange to solve for } \text{T}':$ \\ $(\text{T}')^2 = 2\text{T}^2$ \\ $\text{Take the square root of both sides:}$ \\ $\text{T}' = \sqrt{2\text{T}^2}$ \\ $\text{T}' = \sqrt{2}\text{T}$ \\ $\text{Therefore, the correct answer is option (2).}$