Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)

The problem states that the initial temperature and volume of the gas are T and V, respectively.

According to the problem, VP^3 = constant. Since we need to find the relation between the initial and final temperatures, we can express this equation as follows: \(V_1P_1^3 = V_2P_2^3\) Where \(V_1\) and \(P_1\) are the initial volume and pressure, and \(V_2\) and \(P_2\) are the final volume and pressure.

Since \(V_1P_1^3 = V_2P_2^3\), we know that the product of volume and the cube of pressure remains constant during the expansion. We can now express the final volume as 27V, where V is the initial volume: \((V_1P_1^3) = (27V)P_2^3\) We know that the initial volume is V, so we can divide both sides of the equation by V to find the ratio between the cube of the initial pressure and the cube of the final pressure: \(P_1^3 = 27P_2^3\) Now, from the ideal gas law, we know that \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. So for our initial conditions, we can write: \(P_1V_1 = nRT_1\) and for our final conditions: \(P_2(27V_1) = nRT_2\)

Now, we can eliminate n and R from these equations, since they're constant for the gas throughout the process. So we can rewrite our equation like this: \(\frac{P_1V_1}{T_1} = \frac{P_2(27V_1)}{T_2}\) Now, we know that \(P_1^3 = 27P_2^3\), which means that \(P_2 = \sqrt[3]{\frac{P_1}{27}}\). Plugging this back into the equation above, we get: \(\frac{P_1V_1}{T_1} = \frac{\sqrt[3]{\frac{P_1}{27}}(27V_1)}{T_2}\) Now, we can cancel out \(P_1\) and \(V_1\) from both sides and we're left with: \(\frac{1}{T_1} = \frac{\sqrt[3]{\frac{1}{27}}}{T_2}\) Now, solving for \(T_2\), we get: \(T_2 = \frac{T_1}{\sqrt[3]{\frac{1}{27}}}\) This simplifies to: \(T_2 = 9T_1\) So, the final temperature will be 9 times the initial temperature, which corresponds to the option (A) \(9 \mathrm{~T}\).