A steel container of oxygen has a volume of 20.0 L at 22 C @35 atm. What is the volume @ STP? How many mol of oxygen are in the container?
I would use P1V1/T1 = P2V2/T2 for the first part and PV = nRT for the number of moles.
hi hanna banana!!!
HI!!!! i need some help on this one.. ive been getting a few different answers.. can u help me out with the steps and i can figure it out please
To determine the volume at STP (Standard Temperature and Pressure) and the number of moles of oxygen in the container, 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
First, let's calculate the volume at STP. STP conditions are defined as a pressure of 1 atmosphere (atm) and a temperature of 0 degrees Celsius (°C) or 273.15 Kelvin (K).
To convert the initial temperature of 22°C to Kelvin, we can use the formula:
T(K) = T(°C) + 273.15
T(K) = 22°C + 273.15 = 295.15 K
Now, we can use the following ratio to determine the new volume:
(V1/T1) = (V2/T2)
Where:
V1 = initial volume
T1 = initial temperature
V2 = final volume (at STP)
T2 = STP temperature (273.15 K)
Substituting the values we know:
(V1/295.15 K) = (V2/273.15 K)
V2 = (V1 * 273.15 K) / 295.15 K
V2 = (20.0 L * 273.15 K) / 295.15 K
V2 = 18.4 L (rounded)
Therefore, the volume at STP is approximately 18.4 L.
Next, let's calculate the number of moles of oxygen in the container using the Ideal Gas Law equation:
PV = nRT
Rearranging the equation:
n = (PV) / (RT)
Given:
P = 35 atm (initial pressure)
V = 20.0 L (initial volume)
T = 295.15 K (initial temperature, converted from 22°C)
Let's use the following value for the ideal gas constant (R):
R = 0.0821 L·atm/(mol·K)
Substituting the values:
n = (35 atm * 20.0 L) / (0.0821 L·atm/(mol·K) * 295.15 K)
n = 24.78 mol (rounded)
Therefore, there are approximately 24.78 moles of oxygen in the container.