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If $\mathrm{V}=1$ litre, 10 mole of $\mathrm{H}_2$ and 10 mole of $\mathrm{N}_2$ gas are mixed at temperature $26^{\circ} \mathrm{C}$, then calculate pressure of the gas.
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Verified Answer
The correct answer is:
491 atm
According to ideal gas equation, $\mathrm{pV}=\mathrm{nRT}$.
Given, $\mathrm{V}=1 \mathrm{~L}, \mathrm{n}_1=10$ mole, $\mathrm{n}_2=10$ mole.
$\mathrm{T}=26^{\circ} \mathrm{C}=299 \mathrm{~K} \text {. }$
On substituting the given values in above equation, we get.
$\begin{aligned}
\mathrm{p} & =\frac{20 \mathrm{~mol} \times 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 299 \mathrm{~K}}{1 \mathrm{~L}} \\
& =491 \mathrm{~atm} .
\end{aligned}$
Given, $\mathrm{V}=1 \mathrm{~L}, \mathrm{n}_1=10$ mole, $\mathrm{n}_2=10$ mole.
$\mathrm{T}=26^{\circ} \mathrm{C}=299 \mathrm{~K} \text {. }$
On substituting the given values in above equation, we get.
$\begin{aligned}
\mathrm{p} & =\frac{20 \mathrm{~mol} \times 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 299 \mathrm{~K}}{1 \mathrm{~L}} \\
& =491 \mathrm{~atm} .
\end{aligned}$
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