A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and 300 K?
PV = nRT
You have 20 mols gas. Plug and chug.
To find the volume of the balloon, we can use the ideal gas law equation:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
- Pressure (P) = 1.01 atm
- Temperature (T) = 300 K
- Number of moles of helium gas (n) = 15 moles
- Number of moles of nitrogen gas (n) = 5 moles
First, we need to calculate the total number of moles of gas in the balloon:
Total moles of gas (ntotal) = number of moles of helium gas + number of moles of nitrogen gas
ntotal = 15 moles + 5 moles = 20 moles
Next, we need to calculate the volume (V) using the ideal gas law equation:
V = (n * R * T) / P
The ideal gas constant (R) is a known constant, which is 0.0821 L*atm/(mol*K).
Substituting the given values into the equation, we have:
V = (20 moles * 0.0821 L*atm/(mol*K) * 300 K) / 1.01 atm
Now, we can calculate the value of V:
V = (20 * 0.0821 * 300) / 1.01
V = 4926 / 1.01
V ≈ 4873.27 L
Therefore, the volume of the balloon at 1.01 atm and 300 K is approximately 4873.27 liters.