33. What mass of helium in grams is required to fill 15.0 L balloon to a pressure of1.1 atm at 25°C?

Answer 1

To solve this problem, we can use the ideal gas law:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

In this case, we know the pressure (P = 1.1 atm), volume (V = 15.0 L), and temperature (T = 25°C = 298 K). We need to find the number of moles (n) of helium gas.

First, let's convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298 K

Now we can rearrange the ideal gas law equation to solve for n:
n = PV / RT

n = (1.1 atm)(15.0 L) / (0.0821 L·atm/mol·K)(298 K)
n = 0.18152 mol

Next, we can convert moles to grams using the molar mass of helium, which is about 4.00 g/mol:
mass = n × molar mass
mass = 0.18152 mol × 4.00 g/mol
mass = 0.726 g

Therefore, 0.726 grams of helium is required to fill the 15.0 L balloon to a pressure of 1.1 atm at 25°C.