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Standard XII

Chemistry

Ideal Gas Equation

A 100ml ves...

Question

A $100$ ml vessel containing $O_{2} (g)$ at $1.0$ atm and $400$ K is connected to a $300$ ml vessel containing NO(g) at $1.5$ atm and $400$ K by means of a narrow tube of negligible volume where gases react to form $N O_{2}$ . Final pressure of mixture will be:

1.125

atm

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0.125

atm

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1

atm

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1.5

atm

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Solution

The correct option is A $1.125$ atm
Assume that gases react as ideal gas
$N O + \frac{1}{2} O_{2} ⟶ N O_{2}$
1 mole of $N O$ and $N O_{2}$ and $\frac{1}{2}$ mole of $O_{2}$
$\frac{P_{t o t a l} \times V_{t o t a l}}{m o l e o f N O_{2}} = \frac{P_{1} V_{1}}{n_{1}} + \frac{P_{2} V_{2}}{n_{2}}$

$\frac{P_{t o t a l} \times 400}{1} = \frac{1 \times 300}{1} + \frac{1 \times 100}{\frac{1}{2}}$
$P_{t o t a l} = 1.125 a t m$

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Similar questions

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A 100ml vessel containing O2(g) at 1.0 atm and 400K is connected to a 300ml vessel containing NO(g) at 1.5 atm and 400K by means of a narrow tube of negligible volume where gases react to form NO2. Final pressure of mixture will be:

The correct option is A 1.125 atmAssume that gases react as ideal gasNO+12O2⟶NO21 mole of NO and NO2 and 12 mole of O2Ptotal×Vtotalmole of NO2=P1V1n1+P2V2n2Ptotal×4001=1×3001+1×10012Ptotal=1.125atm

A $100$ ml vessel containing $O_{2} (g)$ at $1.0$ atm and $400$ K is connected to a $300$ ml vessel containing NO(g) at $1.5$ atm and $400$ K by means of a narrow tube of negligible volume where gases react to form $N O_{2}$ . Final pressure of mixture will be: