Nitrogen gas is kept in an open beaker at 273 K and 1 atm pressure. If the pressure of the surrounding suddenly falls to 0.5 atm and the temperature increases to 546 K, then the percentage of niitroge

Question

Nitrogen gas is kept in an open beaker at 273 K and 1 atm pressure. If the pressure of the surrounding suddenly falls to 0.5 atm and the temperature increases to 546 K, then the percentage of niitrogen remaining in the beaker is mn % of the initial amount. Then the value of m+n is:

TestHub · Accepted Answer

Let the initial number of moles of nitrogen gas in the beaker $= n_{1}$

Now, total volume of gas in the beaker will remain constant.
Then, $n \propto \frac{P}{T}$

Therefore,

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{P_{1}}{P_{2}} \times \frac{T_{1}}{T_{2}}$

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{0.5} \times \frac{546}{273} = 4$

Since, the final moles of nitrogen in the beaker is one-fourth of the initial amount in $25$ % of the initial amount.
Therefore, $m = 2, n = 5$ and $m + n = 7$

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