What pressure is exerted by 0.344 mol of $N_{2}$ $N_2$ in a 9.73 L steel container at 105.4°C?

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Answer 1 · 2017-05-15T08:16:22

This is a clear application of the Ideal Gas equation..........I get approx. $1 \cdot a t m$ $1*atm$ .............

$P=(nRT)/V=(0.344*cancel(mol)xx0.0821*(cancelL*atm)/(cancelK*cancel(mol))xx378.5*cancelK)/(9.73*cancelL)$

$~=1*atm$

What pressure is exerted by 0.344 mol of N_2N2N_2 in a 9.73 L steel container at 105.4°C?