The pressure and temperature of an ideal gas in a closed vessel are $720 \mathrm{kPa}\( and \)40^{\circ} \mathrm{C}\( respectively. If \)(1 / 4)^{\text {th }}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to \(353^{\circ} \mathrm{C}\), final pressure of the gas is (A) \(1440 \mathrm{kPa}\) (B) \(540 \mathrm{kPa}\) (C) \(1080 \mathrm{kPa}\) (D) \(720 \mathrm{kPa}\)

To work with the ideal gas law equation, we need to convert all temperatures to Kelvin. To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature. Initial temperature T1: \(T1 = 40^{\circ} \mathrm{C} + 273.15 = 313.15 K\) Final temperature T2: \(T2 = 353^{\circ} \mathrm{C} + 273.15 = 626.15 K\)

First, we use the initial conditions and the ideal gas law equation to determine the initial number of moles (n1). Let V be the volume of the gas in the vessel. Then we have: \(P_1V = n_1RT_1\) Since we're given the initial pressure P1 and the initial temperature T1, we can find n1 as: \(n_1 = \frac{P_1V}{RT_1}\)

A fraction \(\frac{1}{4}\) of the gas is released from the vessel; now, let's find the number of moles after the release of gas (n2). \(n_2 = n_1 - \frac{1}{4} n_1\) \(n_2 = \frac{3}{4} n_1\)

Now we have the remaining moles n2 and the final temperature T2, we can use the ideal gas law again to solve for the final pressure P2. \(P_2V = n_2RT_2\) \(P_2 = \frac{n_2RT_2}{V}\) Now, substitute n2 with the expression found in step 3: \(P_2 = \frac{\frac{3}{4}n_1RT_2}{V}\) Now, we know from step 2: \(n_1 = \frac{P_1V}{RT_1}\) Substituting this expression into the final pressure equation: \(P_2 = \frac{\frac{3}{4}(\frac{P_1V}{RT_1})RT_2}{V}\) Now, simplify the equation: \(P_2 = \frac{3}{4} \frac{P_1T_2}{T_1}\) Finally, put all the given values and found values in the equation to find the final pressure: \(P_2 = \frac{3}{4} \frac{(720 \mathrm{kPa})(626.15 K)}{(313.15 K)}\) \(P_2 = 1080 \mathrm{kPa}\) So, the correct answer is (C) \(1080 \mathrm{kPa}\).