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?
P1V1/T1=P2V2/T2 solve for V2
Now, now many moles...
PV=nRT solve for n OR, when you solve for volume at stp, divide by 22.4liters
i got the answer, 2.37L..and 28.90 mol.. but that doesnt seem right to me, wouldnt it be the other way around?
P1V1/T1= 2.37 L
n=PV/RT= 28.90 mol
or should they be switched?
To find the volume of the steel container 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 solve for the volume @ STP.
STP conditions are defined as:
- Temperature (T) of 0 degrees Celsius or 273.15 Kelvin
- Pressure (P) of 1 atmosphere or 1.01325 bar
We are given:
- T1 = 22 degrees Celsius (Convert to Kelvin: T1 = 22 + 273.15 = 295.15 K)
- P1 = 35 atm
- V1 = 20.0 L
- P2 = 1 atm (STP pressure)
Using the relationship between two sets of conditions:
(P1 * V1) / T1 = (P2 * V2) / T2
Substituting in the known values:
(35 atm * 20.0 L) / 295.15 K = (1 atm * V2) / 273.15 K
Simplifying the equation:
(700 atm * L) / 295.15 K = V2 / 273.15 K
Cross multiplying:
V2 = (700 atm * L * 273.15 K) / 295.15 K
= 645.06 L
Therefore, the volume of the steel container at STP is 645.06 L.
Next, let's determine the number of moles of oxygen in the container.
Rearranging the ideal gas law equation:
n = (PV) / (RT)
We know:
- P = 35 atm
- V = 20.0 L
- T = 295.15 K
Substituting the values:
n = (35 atm * 20.0 L) / (0.0821 atm·L·mol^(-1)·K^(-1) * 295.15 K)
≈ 2.399 mol
Therefore, there are approximately 2.399 moles of oxygen in the container.