Question

Assuming ideal behavior, how many moles of argon would you need to fill a 13.0×12.0×10.0 ft room? Assume atmospheric pressure of 1.00 atm , a room temperature of 20.0 ∘C , and 28.2 L/ft3 . Express your answer with the appropriate units.

Answer 1

To fill a 13.0×12.0×10.0 ft room with argon at 1 atm and 20°C, you would need approximately 1826 moles of argon, calculated using the ideal gas law and the conversion of room dimensions to liters.

Step-by-step explanation:

To determine the number of moles of argon needed to fill a room, we can use the ideal gas law, 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 in Kelvin. First, we need to convert the temperature to Kelvin (20.0 °C = 293.15 K) and the dimensions of the room to cubic feet to liters, then calculate the volume of the room.

The room has dimensions of 13.0 × 12.0 × 10.0 ft, so its volume in cubic feet is 13.0 × 12.0 × 10.0 = 1560 cubic feet. Using the conversion factor 28.2 L/ft³, we can convert this to liters: 1560 ft³ × 28.2 L/ft³ = 44012 L.

Since the pressure is 1.00 atm and the temperature is 293.15 K, we can rearrange the ideal gas law to solve for the number of moles: n = PV / RT. Using the known values, including R = 0.0821 L·atm/mol·K, we calculate the number of moles of argon required:

n = (1.00 atm × 44012 L) / (0.0821 L·atm/mol·K × 293.15 K) ≈ 1826 moles of argon.

Therefore, to fill the room with argon at the given conditions, we would need approximately 1826 moles of argon.

Answer 2

To fill a room of 13.0x12.0x10.0 ft at 1.00 atm and 20.0 °C, 1837 moles of argon are needed.

Step-by-step explanation:

To calculate how many moles of argon are needed to fill a room of 13.0×12.0×10.0 ft at 1.00 atm and 20.0 °C, we first need to convert the room dimensions into liters using the conversion factor 28.2 L/ft³.

Volume in liters = 13.0 ft × 12.0 ft × 10.0 ft × 28.2 L/ft³ = 44112.0 L

Using the Ideal Gas Law (PV=nRT), where P is pressure, V is volume, n is number of moles of the gas, R is the ideal gas constant (0.08206 L·atm/K·mol), and T is temperature in Kelvin.

First, we convert the temperature to Kelvin: T = 20.0 °C + 273.15 = 293.15 K

Now, solve for n (number of moles): n = (P·V) / (R·T) = (1.00 atm · 44112.0 L) / (0.08206 L·atm/K·mol · 293.15 K)

Calculating this out, we get:

n = 1837 moles of argon