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Standard XII

Chemistry

Boyle's Law

A 0.5 dm3...

Question

A $0.5$ $d m^{3}$ flask contains gas 'A' and $1$ $d m^{3}$ flask contains gas 'B' at the same temperature. If density of A $= 3.0$ gm $d m^{- 3}$ and that of B $= 1.5$ gm $d m^{- 3}$ and the molar mass of A $= 1 / 2$ of B, then the ratio of pressure exerted by gases is:

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Solution

Let $P_{A}$ and $P_{B}$ be the pressure exerted by gases $A$ and $B$ respectively.
$∴$ $\frac{P_{A}}{P_{B}} = \frac{\frac{n_{A} R T_{A}}{V_{A}}}{\frac{n_{B} R T_{B}}{V_{B}}} = \frac{M_{B} V_{B}}{M_{A} V_{A}} = \frac{2 M_{A} \times 1}{M_{A} \times 0.5} = 4$

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A 0.5 dm3 flask contains gas 'A' and 1 dm3 flask contains gas 'B' at the same temperature. If density of A=3.0gm dm−3 and that of B=1.5gm dm−3 and the molar mass of A=1/2 of B, then the ratio of pressure exerted by gases is:

Let PA and PB be the pressure exerted by gases A and B respectively.∴ PAPB=nARTAVAnBRTBVB=MBVBMAVA=2MA×1MA×0.5=4

A $0.5$ $d m^{3}$ flask contains gas 'A' and $1$ $d m^{3}$ flask contains gas 'B' at the same temperature. If density of A $= 3.0$ gm $d m^{- 3}$ and that of B $= 1.5$ gm $d m^{- 3}$ and the molar mass of A $= 1 / 2$ of B, then the ratio of pressure exerted by gases is:

Let $P_{A}$ and $P_{B}$ be the pressure exerted by gases $A$ and $B$ respectively.
$∴$ $\frac{P_{A}}{P_{B}} = \frac{\frac{n_{A} R T_{A}}{V_{A}}}{\frac{n_{B} R T_{B}}{V_{B}}} = \frac{M_{B} V_{B}}{M_{A} V_{A}} = \frac{2 M_{A} \times 1}{M_{A} \times 0.5} = 4$