What is the pressure of a gas that originally occupied 4.26 L at a pressure of 36.24 kPa, if the volume is increased by 5.16 L? assume the temperature remains the same.

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Answer 1 · 2018-05-10T00:24:45

Approximately #16.39# kilopascals.

For a constant temperature and number of moles, we can use Boyle's law, which states that,

#Pprop1/V# or #P_1V_1=P_2V_2#

So here, we get:

#4.26 \ "L"*36.24 \ "kPa"=P_2(4.26 \ "L"+5.16 \ "L")#

#154.38 \ "L kPa"=9.42 \ "L"*P_2#

#P_2=(154.38color(red)cancelcolor(black)"L""kPa")/(9.42color(red)cancelcolor(black)"L")#

#~~16.39 \ "kPa"#