4.65 L of nitrogen at standard pressure is compressed into a 0.480 L container. What is the new pressure in mm Hg?

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Answer 1 · 2018-05-07T20:03:47

$P_{2}$ $P_2$ = 7362.5 mmHg

Boyle's Law
$P_{1}$ $P_1$ $V_{1}$ $V_1$ = $P_{2}$ $P_2$ $V_{2}$ $V_2$

(760mmHg)(4.65L) = $P_{2}$ $P_2$ (0.480L)

Divide (760mmHg * 4.65L) by (0.480L) to isolate for $P_{2}$ $P_2$ .

$\frac{760 \cdot 4.65}{0.480} =$ $(760 * 4.65)/(0.480) =$ P_2#

Simplify.

$(\frac{3534}{0.480})$ $(3534/0.480)$ = $P_{2}$ $P_2$

7362.5 mmHg = $P_{2}$ $P_2$