If you have a gas with an initial pressure of 5 atm and it occupies a volume of 10L. If you increase the volume to 13L, what will the new pressure of the gas be?

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Answer 1 · 2018-02-26T04:45:35

$4 L$ $"4 L"$

Use Boyle's Law

$pressure(1) \times volume(1) = pressure(2) \times volume(2)$ $"pressure(1)" xx "volume(1)" = "pressure(2)" xx "volume(2)"$

where

$pressure(1) = 5 atm$ $"pressure(1) = 5 atm"$

$volume(1) = 10 L$ $"volume(1) = 10 L"$

$pressure(2) = 13 atm$ $"pressure(2)" = "13 atm"$

Therefore

$5 atm \times 10 L = 13 atm \times volume(2)$ $"5 atm" xx "10 L" ="13 atm" xx "volume(2)"$

$volume(2) = \frac{5 atm \times 10 L}{13 atm} = 3.85 L = 4 L \to$ $"volume(2)" = ("5 atm" xx "10 L")/"13 atm" = "3.85 L" = "4 L" ->$ one sigfig