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Standard XIII

Chemistry

Dalton's Law of Partial Pressures

The pressure ...

Question

The pressure of $1 g$ of an ideal gas $A$ at $300 K$ is found to be 2 bar. When $2 g$ of another ideal gas $B$ is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Which of the following is correct?
( $m_{A}$ = mass of gas A; $m_{B}$ = mass of gas B)

m_{B} = 4 m_{A}

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m_{A} = 3 m_{B}

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m_{B} = 3 m_{A}

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m_{A} = m_{B}

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Solution

The correct option is A $m_{B} = 4 m_{A}$
Given,
Mass of gas A, $w_{A} = 1 g$
Mass of gas B, $w_{B} = 2 g$
$p_{A} = 2 bar$
$P_{t o t a l} = 3 bar$
$P_{t o t a l} = p_{A} + p_{B}$
$p_{B} = 3 - 2 = 1 bar$
$P V = n R T \Rightarrow P V = \frac{W}{M} R T$
$\frac{p_{A} m_{A}}{w_{A}} = \frac{p_{B} m_{B}}{w_{B}} or \frac{2 \times m_{A}}{1} = \frac{1 \times m_{B}}{2} or \frac{m_{A}}{m_{B}} = \frac{1}{4} or m_{B} = 4 m_{A}$

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Dalton's Law of Partial Pressures

Standard XIII Chemistry

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The pressure of 1 g of an ideal gas A at 300 K is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Which of the following is correct? (mA = mass of gas A; mB = mass of gas B)

The correct option is A mB=4 mAGiven, Mass of gas A, wA=1 g Mass of gas B, wB=2 g pA=2 bar Ptotal=3 bar Ptotal=pA+pB pB=3−2=1 bar PV=nRT⇒PV=WMRT pAmAwA=pBmBwB or 2×mA1=1×mB2 or mAmB=14 or mB=4 mA

The pressure of $1 g$ of an ideal gas $A$ at $300 K$ is found to be 2 bar. When $2 g$ of another ideal gas $B$ is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Which of the following is correct?
( $m_{A}$ = mass of gas A; $m_{B}$ = mass of gas B)