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Dalton's Law of Partial Pressures

In a closed c...

Question

In a closed container 3 moles of $A$ and 2 moles of $B$ with 3 moles of $C$ are present at pressure 40 atm. Find partial pressure of each.

P_{A} = P_{B} = 15 a t m

and

P_{C} = 10 a t m

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P_{A} = P_{C} = 15 a t m

and

P_{B} = 10 a t m

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P_{A} = 15 a t m, P_{B} = 10 a t m

and

P_{C} = 5 a t m

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none of these

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Solution

The correct option is B $P_{A} = P_{C} = 15 a t m$ and $P_{B} = 10 a t m$
Partial pressure of A $= P_{A} = \frac{moles of A}{total moles} \times total pressure$
$= \frac{3}{3 + 3 + 2} \times 40$
$= \frac{3}{8} \times 40$
$P_{A} = 15 a t m$
Potential pressure of B $P_{B} = \frac{2}{8} \times 40$
$= 10 a t m$
Partial pressure of C $= P_{C} = \frac{3}{8} \times 40$
$P_{C} = 15 a t m$

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In a closed container 3 moles of A and 2 moles of B with 3 moles of C are present at pressure 40 atm. Find partial pressure of each.

The correct option is B PA=PC=15 atm and PB=10 atmPartial pressure of A =PA=moles of Atotal moles×total pressure =33+3+2×40 =38×40 PA=15atmPotential pressure of B PB=28×40 =10 atmPartial pressure of C =PC=38×40 PC=15 atm

In a closed container 3 moles of $A$ and 2 moles of $B$ with 3 moles of $C$ are present at pressure 40 atm. Find partial pressure of each.