Question

Question #4d509

Answer 1

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.`

Answer 2

`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!

Answer 3

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 :)