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A vessel has nitrogen gas and water vapour at a total pressure of 1 atm . The partial pressure of water vapour is 0.3 atm. When the contents of this vessel are transformed to another vessel having one third of the capacity of original vessel , completely at the same temperature , the total pressure of the system in the new vessel is:

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

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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)
answered Mar 31, 2014 by sharmaaparna1
 

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