A balloon is inflated to a volume of 7.0 L on a day when the atmospheric pressure is 765 mmHg. The next day, as a storm front arrives, the atmospheric pressure drops to 725 mmHg. Assuming the temperature remains constant, what is the new volume of the balloon, in liters?
the volume is inversely proportional to the ambient pressure
v = 7.0 L * (765 / 725)
To find the new volume of the balloon, we can use the relationship between pressure and volume known as Boyle's Law. Boyle's Law states that at a constant temperature, the product of pressure and volume is constant.
Mathematically, Boyle's Law can be expressed as:
(P1)(V1) = (P2)(V2)
Where P1 and P2 are the initial and final pressures of the balloon, and V1 and V2 are the initial and final volumes of the balloon.
In this case, we know that the initial volume (V1) is 7.0 L and the initial pressure (P1) is 765 mmHg. The final pressure (P2) is given as 725 mmHg. We need to find the final volume (V2).
Let's substitute these values into Boyle's Law and solve for V2:
(765 mmHg)(7.0 L) = (725 mmHg)(V2)
By rearranging the equation, we can isolate V2:
V2 = (765 mmHg)(7.0 L) / 725 mmHg
Now let's calculate the value of V2:
V2 = 5295 mmHg * L / 725 mmHg
V2 ≈ 7.3 L
Therefore, the new volume of the balloon is approximately 7.3 liters.
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is constant.
According to Boyle's Law:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure (in mmHg)
V1 = initial volume (in liters)
P2 = final pressure (in mmHg)
V2 = final volume (in liters)
Given:
P1 = 765 mmHg
V1 = 7.0 L
P2 = 725 mmHg
Now, let's plug these values into the equation and solve for V2:
P1 * V1 = P2 * V2
765 mmHg * 7.0 L = 725 mmHg * V2
5355 mmHg * L = 725 mmHg * V2
Dividing both sides of the equation by 725 mmHg:
5355 mmHg * L / 725 mmHg = 725 mmHg * V2 / 725 mmHg
7.38 L = V2
Therefore, the new volume of the balloon is approximately 7.38 liters.