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Byju's Answer

Standard XII

Chemistry

Boyle's Law

The volumes o...

Question

The volumes of two vessels at same temperature are in theratio of 2:3 . One vessel contains $H_{2}$ and other $N_{2}$ at 600 mm and 900 mm respectively. The final pressure when they are connected together is : (Assume that $N_{2}$ and $H_{2}$ react to form $N H_{3}$ )

680 mm

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760 mm

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780 mm

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800 mm

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Solution

The correct option is A 680 mm
Let volume of vessel be $2 V$ and $3 V$
$^{n} H_{2} = \frac{P V}{R T} = \frac{600 \times 2 V}{R \times T} = \frac{1200 V}{R T}$

$^{n} N_{2} = \frac{P V}{R T} = \frac{2700 V}{R T}$

Total $n = \frac{3900 V}{R T}, P V = n R T$

$V_{t o t a l} = 5 V$ $\Rightarrow P_{f i n a l} = \frac{n R T}{V} = \frac{3900 V}{R T} \times \frac{R T}{5 V} = 680 n m$

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The volumes of two vessels at same temperature are in theratio of 2:3 . One vessel contains H2 and other N2 at 600 mm and 900 mm respectively. The final pressure when they are connected together is : (Assume that N2 and H2 react to form NH3)

The correct option is A 680 mmLet volume of vessel be 2V and 3VnH2=PVRT=600×2VR×T=1200 VRTnN2=PVRT=2700VRTTotal n=3900VRT,PV=nRTVtotal=5V ⇒Pfinal=nRTV=3900VRT×RT5V=680 nm

The volumes of two vessels at same temperature are in theratio of 2:3 . One vessel contains $H_{2}$ and other $N_{2}$ at 600 mm and 900 mm respectively. The final pressure when they are connected together is : (Assume that $N_{2}$ and $H_{2}$ react to form $N H_{3}$ )