A balloon filled with 3.5 L of helium at 20 °C and 3 atm. What will be the new pressure if its volume increases to 17.5 L at constant temperature?

PV = kT.

So, PV/T is constant. In fact, since T is constant as well, PV is constant. You want P so that

17.5P = 3*3.5

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional, assuming that the temperature remains constant. Mathematically, Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

Given:
P₁ = 3 atm
V₁ = 3.5 L
V₂ = 17.5 L
T = Constant

Let's substitute these values into the equation and solve for P₂:

P₁V₁ = P₂V₂

3 atm * 3.5 L = P₂ * 17.5 L

10.5 atm-L = 17.5 P₂

Now divide both sides by 17.5 to solve for P₂:

P₂ = (10.5 atm-L) / 17.5

P₂ ≈ 0.6 atm

Therefore, the new pressure of the balloon would be approximately 0.6 atm when its volume increases to 17.5 L at constant temperature.

To find the new pressure of the balloon when its volume increases to 17.5 L at constant temperature, we can use Boyle's Law. Boyle's Law states that the pressure exerted by a gas is inversely proportional to its volume, as long as the temperature remains constant.

Boyle's Law can be mathematically represented as:

P1 × V1 = P2 × V2

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given:
Initial volume of the balloon (V1) = 3.5 L
Initial pressure of the balloon (P1) = 3 atm
Final volume of the balloon (V2) = 17.5 L (after the volume increases)
Final pressure of the balloon (P2) = ?

Plugging in the values into the equation, we get:

3 atm × 3.5 L = P2 × 17.5 L

To solve for P2, divide both sides of the equation by 17.5 L:

(3 atm × 3.5 L) / 17.5 L = P2

Multiply and divide:

10.5 atm L / 17.5 L = P2

Now, simplify:

10.5 atm / 17.5 = P2

Calculating this value, we find:

P2 ≈ 0.6 atm

Therefore, the new pressure of the balloon will be approximately 0.6 atm when its volume increases to 17.5 L at constant temperature.