Two closed vessels of same volume are joined through a narrow tube...

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: $100$