Answer:
the density of the helium gas would be approximately 0.369 g/L when the balloon is placed in the freezer at -10°C and a pressure of 2.0 atm.
Step-by-step explanation:
To calculate the density of helium gas in the balloon after it is placed in the freezer at -10°C and a pressure of 2.0 atm, we can use the ideal gas law and the relationship between density, molar mass, and molar volume.
First, let's find the initial molar volume of the helium gas using the given conditions:
PV = nRT
Where:
P = pressure = 1.0 atm
V = volume = 0.75 L
n = number of moles = 0.0303 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin
To convert Celsius to Kelvin, we add 273.15:
T = 30°C + 273.15 = 303.15 K
Using the ideal gas law, we can calculate the initial molar volume:
V_initial = (n * R * T) / P
V_initial = (0.0303 mol * 0.0821 L·atm/(mol·K) * 303.15 K) / 1.0 atm
V_initial ≈ 0.754 L
Next, we can calculate the molar mass of helium (He) using the atomic mass of helium:
Molar mass of He = 4.003 g/mol
Now we can calculate the initial density of the helium gas in the balloon:
Initial density = (mass of helium gas) / (volume of helium gas)
Initial density = (0.0303 mol * 4.003 g/mol) / 0.754 L
Initial density ≈ 0.161 g/L
Now let's find the final density of the helium gas when the balloon is placed in the freezer at -10°C and a pressure of 2.0 atm.
We will use the ideal gas law again with the new conditions:
P_final = 2.0 atm
T_final = -10°C + 273.15 = 263.15 K (converted to Kelvin)
To find the final molar volume, we rearrange the ideal gas law equation:
V_final = (n * R * T_final) / P_final
V_final = (0.0303 mol * 0.0821 L·atm/(mol·K) * 263.15 K) / 2.0 atm
V_final ≈ 0.328 L
Finally, we can calculate the final density of the helium gas:
Final density = (mass of helium gas) / (volume of helium gas)
Final density = (0.0303 mol * 4.003 g/mol) / 0.328 L
Final density ≈ 0.369 g/L