STP convention
STP stands for standard temperature and pressure
These standard conditions are:
P = 1 atm
T = 0EC or 273.15 K
You may see standard conditions referred
to in either a straight use of the ideal gas
law or a parametric use.
The straight subsititution is quite easy.
First a quick derivation:
Question:
What is the volume of exactly 1 mole of
ideal gas at STP?
P = 1 atm*
T = 273.15 K
n = 1 mol*
* an exact value
Question:
What is the volume of exactly 1 mole of
ideal gas at STP?
P = 1 atm*
T = 273.15 K
n = 1 mol*
Substituting into PV = nRT:
(1 atm)V = (1 mol)(0.08206 L atm K -1 mol-1)(273.15 K)
* an exact value
Question:
What is the volume of exactly 1 mole of
ideal gas at STP?
P = 1 atm*
T = 273.15 K
n = 1 mol*
(1 atm)V = (1 mol)(0.08206 L atm K -1 mol-1)(273.15 K)
and solving:
V = 22.41 L
* an exact value
Thus, 1 mole of ideal gas at STP has a
volume of 22.41 L or one could say the
molar volume at STP is 22.41 L:
VSTP = 22.41 L mol-1
Let’s take a few examples where this could
be used.
VSTP = 22.41 L mol-1
or an equation could include this:
V = VSTPn
How many moles of gas are there in a 5.0 L
volume at STP?
Substituting in:
5.0 L = n (22.41 L mol-1)
n = 0.22 L
VSTP = 22.41 L mol-1
or an equation could include this:
V = VSTPn
How many grams of hydrogen is in 6800
liters at STP?
First obtain the number of moles by
substituting:
6800 L = (22.41 L mol-1)n
n = 303.4 mol
VSTP = 22.41 L mol-1
or an equation could include this:
V = VSTPn
How many grams of hydrogen is in 6800
liters at STP?
n = 303.4 mol
Now convert the moles of hydrogen to
grams using m = Mn:
-1
m = (2.016 g mol )(303.4 mol)
m = 611.7 g
Example of using STP in a problem:
A gas is contained at 373 K in a 4.0 L
volume at 1.00 atm. What is its volume at
STP?
V
P
V
P
1
1
2
2
Here P and n do not change so:
=
n1T1
n2T2
V1 V2
=
and:
T1
T2
Example of using STP in a problem:
A gas is contained at 373 K in a 4.0 L
volume at 1.00 atm. What is its volume at
STP?
P
V
P
V
1
1
2
2
Here P and n do not change so:
=
n1T1
n2T2
V1 V2
=
and:
T1
T2
The values are:
V1 = 4.0 L
V2 = ?
T1 = 373 K
T2 = 273 K (from STP)
Example of using STP in a problem:
A gas is contained at 373 K in a 4.0 L
volume at 1.00 atm. What is its volume at
STP?
V1 V2
=
T1
T2
V1 = 4.0 L
V2 = ?
T1 = 373 K
T2 = 273 K (from STP)
substuting: 4.0 L
V2
=
373 K 273 K
and evaluating:
V2 = 2.9 L
THE END
Use the following keys if you do not want to exit.
control home - starts the show over
right arrow (º) advances to the next slide
left arrow (») goes back to the previous slide
try them.