Answer:
The volume of that ballon would be approximately [tex]1.20 \times 10^2\; \rm L[/tex] at a pressure of [tex]0.250\; \rm atm[/tex].
Explanation:
Let [tex]V_1[/tex] and [tex]P_1[/tex] be the volume and pressure of the gas before the change.
By Boyle's Law, the volume of an ideal gas is inversely proportional to its pressure. In other words, if [tex]V_2[/tex] and [tex]P_2[/tex] represent the volume after the change, then [tex]V_1 \cdot P_1 = V_2 \cdot P_2[/tex].
In this case,
The value of [tex]V_2[/tex] needs to be found. Rearrange the equation [tex]V_1 \cdot P_1 = V_2 \cdot P_2[/tex]:
[tex]\begin{aligned}& V_1\cdot P_1 = V_2 \cdot P_2 \\ & \implies V_2 = \frac{V_1 \cdot P_1}{P_2} = \frac{P_1}{P_2} \cdot V_1\end{aligned}[/tex].
Therefore,
[tex]\begin{aligned}\displaystyle V_2 &= \frac{P_1}{P_2}\cdot V_1 \\ &= \frac{1.00\; \rm atm}{0.250\; \rm atm} \times 30.0\; \rm L = 1.20 \times 10^2 \; \rm L\end{aligned}[/tex].
