An experimental chamber has a volume of 60 L. How many moles of oxygen gas will be required to fill the chamber at STP?
PV = nRT
2.7 Moles because
60L/22.4L gives you 2.7 moles
The formula for STP for gasses is: 1mole/22.4L
Therefore, 60Liters(1mole/22.4L)=2.68 moles or 2.7 moles.
To determine the number of moles of oxygen gas required to fill the chamber at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
At STP, the standard conditions are:
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/(mol·K)
Given:
V = 60 L
Using the ideal gas law equation, we can rewrite it to solve for the number of moles (n):
n = PV / RT
Substituting the known values:
n = (1 atm) * (60 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = 2.26 mol
Therefore, approximately 2.26 moles of oxygen gas will be required to fill the chamber at STP.
To calculate the number of moles of oxygen gas required to fill the chamber at STP (Standard Temperature and Pressure), we need to use the ideal gas law. The ideal gas law equation is:
PV = nRT
where:
P = pressure (in units of pressure, such as atmospheres)
V = volume (in units of volume, such as liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in units of temperature, such as Kelvin)
At STP, the temperature is 273 K, and the pressure is 1 atm. Since we are given the volume of the chamber as 60 L, we can plug in these values into the ideal gas law equation and solve for n:
(1 atm)(60 L) = n(0.0821 L·atm/(mol·K))(273 K)
Now, we can solve for n:
60 atm·L = 22.4343 n
Divide both sides of the equation by 22.4343:
n = 60 atm·L / 22.4343
n ≈ 2.674 moles
Therefore, approximately 2.674 moles of oxygen gas will be required to fill the chamber at STP.