If #9/4 L# of a gas at room temperature exerts a pressure of #15 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/3 L#?

Question

1337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-23T00:26:43

#20.25# kilopascals

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

#Pprop1/V# or #P_1V_1=P_2V_2#

where #P_1,P_2# are the initial and final pressure of the gas, and #V_1,V_2# are the initial and final volume of the gas.

So, we get:

#9/4 \ "L"*15 \ "kPa"=P_2*5/3 \ "L"#

#:.P_2=(9/4color(red)cancelcolor(black)"L"*15 \ "kPa")/(5/3color(red)cancelcolor(black)"L")#

#=81/4 \ "kPa"#

#=20.25 \ "kPa"#