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Standard IX

Chemistry

Elements

Two flask A ...

Question

Two flask A & B have capacity of 1 litre and 2 litre respectively. Each of them contain 1 mole of a gas. The temperature of a flask are so adjusted that average speed of molecules in "A" is twice than in "B" & pressure in flask "A" is x times of that in "B". Then value of x is-

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Solution

$\begin{matrix} \begin{matrix} 1 m o l A A (1 L) \end{matrix} \begin{matrix} 1 m o l B B (2 L) \end{matrix} & \frac{V_{a v} (a)}{V_{a v} (b)} = \frac{\sqrt{8 R T / π M a}}{\sqrt{8 R T / π M b}} \Rightarrow \sqrt{\frac{M b}{M a}} \frac{2}{1} = \sqrt{\frac{M b}{M a}}; \frac{M b}{M a} = 4; M b = 4 M a p_{A} = \frac{n_{A} R T}{V_{A}} P_{B} = \frac{n_{B} R T}{V_{B}} & \frac{P_{A}}{P_{B}} = \frac{n_{A} v_{B}}{n_{B} v_{A}} \Rightarrow \frac{1 \times 2}{1 \times 1} = \frac{2}{1} P_{A} = 2 P_{B}, x = 2 \end{matrix}$

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Two flask A & B have capacity of 1 litre and 2 litre respectively. Each of them contain 1 mole of a gas. The temperature of a flask are so adjusted that average speed of molecules in "A" is twice than in "B" & pressure in flask "A" is x times of that in "B". Then value of x is-

1molAA(1L)1molBB(2L)Vav(a)Vav(b)=√8RT/πMa√8RT/πMb⇒√MbMa21=√MbMa;MbMa=4;Mb=4MapA=nARTVAPB=nBRTVBPAPB=nAvBnBvA⇒1×21×1=21PA=2PB,x=2