calculate the pressure in atm of a balloon filled with 7.5g of helium gas if the baloons volume decreases to 7.5L suppose the balloons pressure and volume are originally 799 mmHg and 15L

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Calculate the pressure in atm of a balloon filled with 7.50g of helium gas if the balloon's volume decreases to 7.50L. Suppose the balloon's pressure and volume are originally 799 mmHg and 15.0L.

P1V1 = P2V2

To calculate the pressure in atm of the balloon filled with 7.5g of helium gas, you can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's convert the original pressure from mmHg to atm by dividing it by 760 mmHg/atm:

799 mmHg / 760 mmHg/atm = 1.050 atm

Next, we need to determine the number of moles of helium gas. To do this, we'll use the molar mass of helium (4.0026 g/mol) and divide the mass of helium gas by its molar mass:

n = m/M
n = 7.5 g / 4.0026 g/mol ≈ 1.875 mol

Now that we have the number of moles, we can rearrange the ideal gas law equation to solve for the new pressure (P2):

P1V1 = nRT1
P2V2 = nRT2

We can assume that the amount of helium gas (number of moles) and the temperature remain constant. Rearranging the equation, we get:

P2 = (P1V1) / V2

Substituting the given values:

P2 = (1.050 atm * 15 L) / 7.5 L
P2 = 2.1 atm

Therefore, the pressure in atm of the balloon filled with 7.5g of helium gas, when the balloon's volume decreases to 7.5L, is approximately 2.1 atm.