Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure \(90\) kPa and temperature \(400\) K. Keeping the temperature of one vessel constant at \(400\) K the second vessel temperature is raised to \(500\) K. The final pressure in the vessels is _______ kPa.

Let the volume of each vessel be \(V\).

Initially, both vessels are at pressure \(P_1 = 90\) kPa and temperature \(T_1 = 400\) K.

Total number of moles initially is given by:

\(n_i = \dfrac{P_1 V}{R T_1} + \dfrac{P_1 V}{R T_1} = \dfrac{2 P_1 V}{R T_1}\)

Finally, the temperature of the first vessel is \(T_1 = 400\) K and the second vessel is \(T_2 = 500\) K. Let the final pressure be \(P_f\).

Total number of moles finally is given by:

\(n_f = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}\)

Since the total number of moles remains constant, \(n_i = n_f\):

\(\dfrac{2 P_1 V}{R T_1} = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}\)

Substituting the given values:

\(\dfrac{2 \times 90}{400} = \dfrac{P_f}{400} + \dfrac{P_f}{500}\)

\(\dfrac{180}{400} = P_f \left( \dfrac{1}{400} + \dfrac{1}{500} \right)\)

\(\dfrac{180}{400} = P_f \left( \dfrac{5 + 4}{2000} \right)\)

\(\dfrac{180}{400} = P_f \left( \dfrac{9}{2000} \right)\)

\(P_f = \dfrac{180 \times 2000}{400 \times 9}\)

\(P_f = 20 \times 5 = 100\) kPa

Answer: \(100\)