Question

a balloon contains 9.30 l of he. the pressure is reduced to 4.00 atm, and the balloon expands to occupy a volume of 25.2 l. what was the initial pressure exerted on the balloon (in atm)?

Answer 1

By applying Boyle's Law, which states that pressure times volume is constant for a given amount of gas at constant temperature, we calculate the initial pressure on a helium balloon to be approximately 10.86 atm when the pressure is reduced to 4.00 atm and the volume increases to 25.2 L.

Step-by-step explanation:

To determine the initial pressure exerted on the balloon, we can use the ideal gas law in the form of Boyle's Law, since the temperature and the amount of helium remain constant. Boyle's Law states that the pressure of a gas times its volume is constant (P1V1 = P2V2). In this case, after the pressure is reduced to 4.00 atm, the balloon expands to occupy a volume of 25.2 L. We are given that the initial volume of the balloon is 9.30 L and need to find the initial pressure (P1).

Using Boyle's Law:
P1 × 9.30 L = 4.00 atm × 25.2 L

Now, solving for P1 gives:
P1 = (4.00 atm × 25.2 L) / 9.30 L
P1 = 10.86 atm (rounded to two decimal places)

Therefore, the initial pressure exerted on the balloon was approximately 10.86 atm.