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Standard XII

Chemistry

Dalton's Law of Partial Pressures

Pressure of 1...

Question

Pressure of $1 g$ of an ideal gas $A$ at $27^{\circ} C$ 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. Find a relationship between their molecular masses.

M_{A} = 4 M_{B}

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M_{B} = 4 M_{A}

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M_{A} = 2 M_{B}

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M_{B} = 2 M_{A}

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Solution

The correct option is B $M_{B} = 4 M_{A}$
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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Pressure of 1 g of an ideal gas A at 27 °C 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. Find a relationship between their molecular masses.

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Dalton's Law of Partial Pressures

Standard XII Chemistry

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Pressure of 1 g of an ideal gas A at 27∘C 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. Find a relationship between their molecular masses.

The correct option is B MB=4MAGiven, 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=4MA

Pressure of $1 g$ of an ideal gas $A$ at $27^{\circ} C$ 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. Find a relationship between their molecular masses.