$(a)\;3.0 atm\qquad(b)\;1 atm\qquad(c)\;3.33 atm\qquad(d)\;2.4 atm$

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$P_{N_2} + P_{H_2O} = 1 atm$

$\therefore P_{H_2O} = 0.3\;atm$

$\therefore P_{N_2}$ = 0.7 atm in 1 vessel .

Now new pressure of $N_2$ in another vessel may be calculated as

$P_1V_1 = P_2V_2$

$\therefore 0.7 \times V_1 = P_2\times \large\frac{V_1}{3}$atm

$\therefore P_{N_2}$ = 2.1 atm

$\Rightarrow P_{H_2O} = 0.3$atm

$P_T = 2.4$ atm

$P_{H_2O}$ remains constant at constant temperature

Hence answer is (D)

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