# Question 4d509

Jun 19, 2017

We use old Boyle's Law..........${P}_{\text{initial}} = 10 \cdot a t m .$

#### Explanation:

........which holds that $P V = k$, and thus ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.

And thus ${P}_{1} = \frac{{P}_{2} {V}_{2}}{V} _ 1 = \frac{5 \cdot a t m \times 500 \cdot m L}{250 \cdot m L} = 10 \cdot a t m .$

Jun 19, 2017

$10$ $\text{atm}$

#### Explanation:

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

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

We're asked to find the original pressure exerted by the gas, so let's rearrange to solve for ${P}_{1}$:

${P}_{1} = \frac{{P}_{2} {V}_{2}}{{V}_{1}}$

sfcolor(red)("Our known quantities":

• ${V}_{1} = 250$ $\text{mL}$ (original volume)

• ${P}_{2} = 5$ $\text{atm}$ (final pressure)

• ${V}_{2} = 500$ $\text{mL}$ (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")

$10 \times 250 = \textcolor{g r e e n}{2500}$ (${P}_{1} \times {V}_{1}$)

5 xx 500 = color(green)(2500# (${P}_{2} \times {V}_{2}$)

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

$\therefore$ $p 1 v 1$ = $p 2 v 2$

$p 1 \times 250 = 5 \times 500$

$p 1 = \frac{5 \times 500}{250}$

$p 1 = 10 a t m$

$\to$ The volume of the gas before expansion was hence 10 atmospheres.

Hope that helped :)