Question #4d509

3 Answers
Jun 19, 2017

We use old Boyle's Law..........#P_"initial"=10*atm.#

Explanation:

........which holds that #PV=k#, and thus #P_1V_1=P_2V_2#.

And thus #P_1=(P_2V_2)/V_1=(5*atmxx500*mL)/(250*mL)=10*atm.#

Jun 19, 2017

#10# #"atm"#

Explanation:

To solve this, we can use the pressure-volume relationship of gases, illustrated by Boyle's law:

#P_1V_1 = P_2V_2#

We're asked to find the original pressure exerted by the gas, so let's rearrange to solve for #P_1#:

#P_1 = (P_2V_2)/(V_1)#

#sfcolor(red)("Our known quantities"#:

  • #V_1 = 250# #"mL"# (original volume)

  • #P_2 = 5# #"atm"# (final pressure)

  • #V_2 = 500# #"mL"# (final volume)

Let's plug in these values into the equation to find the original pressure:

#P_1 = (P_2V_2)/(V_1) = ((5color(white)(l)"atm")(500cancel("mL")))/((250cancel("mL"))) = color(blue)(10# #color(blue)("atm"#

Thus, the original pressure of the gas was #color(blue)(10# #sfcolor(blue)("atmospheres"#.

We can check this answer by plugging all the values into the original equation:

#P_1V_1 = P_2V_2 = (color(blue)(10)color(white)(l)color(blue)("atm"))(250color(white)(l)"mL") = (5color(white)(l)"atm")(500color(white)(l)"mL")#

#10 xx 250 = color(green)(2500)# (#P_1xxV_1#)

#5 xx 500 = color(green)(2500# (#P_2xxV_2#)

Our answer is furthermore correct!

Jun 19, 2017

As per Boyle's law the answer is

Explanation:

Boyle's law states that if the pressure of a gas changes from p1 to p2 then the volume would also change from v1 to v2 at a constant temperature. The pressure is also inversely proportional to the volume

#:.# #p1v1# = #p2 v2#

#p1 xx 250 = 5 xx 500#

#p1 = (5 xx 500)/250#

#p1 = 10 atm#

#-># The volume of the gas before expansion was hence 10 atmospheres.

Hope that helped :)