1L flask contains nitrogen along with a drop or two of water at $40^{\circ} C$ , the total pressure being 760 torr. If all the conents are transferred to another flask of 0.5 L at the same temperature, the pressure set up in the second flask will be (aqueous tension at $40^{\circ} C = 55 t o r r)$

1410 torr

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1465 torr

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1520 torr

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None of these

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Solution

The correct option is B
1465 torr

$P_{1} (d r y) \times V_{1} = P_{2} (d r y) \times V_{2}; P_{1} = P_{w e t} - a q u e o u s t e n s i o n = 760 - 55 = 705 t o r r$
$V_{2} = 0.5 L,$ Hence, $P_{2}$ (dry gas) $= \frac{705 \times 1}{0.5} = 1410 t o r r$
$P_{2}$ (Wet gas) $= 1410 + 55 = 1465 t o r r$

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1 L flask contains nitrogen along with a drop or two of water at $40^{\circ} C$ , the total pressure is 760 torr. If all the contents are transferred to another flask of 0.5 L at the same temperature, the pressure set up in the second flask will be (aqueous tension at $40^{\circ} C = 55 t o r r)$

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1L flask contains nitrogen along with a drop or two of water at 40∘C, the total pressure being 760 torr. If all the conents are transferred to another flask of 0.5 L at the same temperature, the pressure set up in the second flask will be (aqueous tension at 40∘C=55 torr)

The correct option is B 1465 torr P1(dry)×V1=P2(dry)×V2;P1=Pwet−aqueous tension=760−55=705 torr V2=0.5L, Hence, P2 (dry gas) =705×10.5=1410 torr P2 (Wet gas) =1410+55=1465 torr

1L flask contains nitrogen along with a drop or two of water at $40^{\circ} C$ , the total pressure being 760 torr. If all the conents are transferred to another flask of 0.5 L at the same temperature, the pressure set up in the second flask will be (aqueous tension at $40^{\circ} C = 55 t o r r)$

The correct option is B
1465 torr

$P_{1} (d r y) \times V_{1} = P_{2} (d r y) \times V_{2}; P_{1} = P_{w e t} - a q u e o u s t e n s i o n = 760 - 55 = 705 t o r r$
$V_{2} = 0.5 L,$ Hence, $P_{2}$ (dry gas) $= \frac{705 \times 1}{0.5} = 1410 t o r r$
$P_{2}$ (Wet gas) $= 1410 + 55 = 1465 t o r r$