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
14 years ago
14 years ago
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.
14 years ago
P1V1 = P2V2
11 months ago
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.