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Standard XII

Chemistry

Heat Capacity

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 $m_{B} = 4 m_{A}$
$P_{A} = \frac{n_{A} R T}{V} = \frac{1 \times R T}{m_{A} V} = 2 bar$

$P_{A} + P_{B} = 3 bar$

$∴ P_{B} = 3 - 2 = 1 bar$

$P_{B} = \frac{2 R T}{m_{B} V} = 1 \Rightarrow \frac{P_{B}}{P_{A}} = \frac{2 m_{A}}{m_{B}} = \frac{1}{2}$

Thus, $m_{B} = 4 m_{A}$

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Q. The pressure of

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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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Heat Capacity

Standard XII 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=4mAPA=nARTV=1×RTmAV=2 bar PA+PB=3 bar ∴ PB=3−2=1 bar PB=2RTmBV=1⇒PBPA=2mAmB=12 Thus, mB=4mA

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 $)$

The correct option is A $m_{B} = 4 m_{A}$
$P_{A} = \frac{n_{A} R T}{V} = \frac{1 \times R T}{m_{A} V} = 2 bar$

$P_{A} + P_{B} = 3 bar$

$∴ P_{B} = 3 - 2 = 1 bar$

$P_{B} = \frac{2 R T}{m_{B} V} = 1 \Rightarrow \frac{P_{B}}{P_{A}} = \frac{2 m_{A}}{m_{B}} = \frac{1}{2}$

Thus, $m_{B} = 4 m_{A}$