# What pressure is exerted by 0.344 mol of N_2 in a 9.73 L steel container at 105.4°C?

This is a clear application of the Ideal Gas equation..........I get approx. $1 \cdot a t m$.............
$P = \frac{n R T}{V} = \frac{0.344 \cdot \cancel{m o l} \times 0.0821 \cdot \frac{\cancel{L} \cdot a t m}{\cancel{K} \cdot \cancel{m o l}} \times 378.5 \cdot \cancel{K}}{9.73 \cdot \cancel{L}}$
$\cong 1 \cdot a t m$