# Standard Temperature and Pressure (STP) ```The Combined and Ideal
Gas Laws
CP Chemistry
The Combined Gas Law
P1V1 = P2V2
T1
T2

Example: A balloon with Helium has a volume of 50.L at
25oC and a pressure of 1.08 atm. What volume will it
have at 0.855 atm and 10oC?
**Convert Temps to Kelvin
T1 = 25 + 273 = 298K
T2 = 10 + 273 = 283K
Plug in and Solve:
(1.08atm)(50L) = (0.855atm)V2
298K
283 K
(1.08)(50)(283) = (0.855)(298) V2
V2 = 60. Liters
Standard Temperature and
Pressure (STP)
to a Temperature of 0oC , and a
pressure of 1 atm.
 Refers
 Keep
in mind the STP values in other
units:
Standard Temp = 0oC or 273 Kelvin
Standard Pressure = 1atm = 760 mmHg
= 760 torr
= 101.325kPa
Ideal Gas

An imaginary gas that fits all the assumptions of
the kinetic molecular theory of gases.

In reality, the noble gases (He, Neon, etc) are
the only gases that most closely behave like an
ideal gas.

All gases behave like an ideal gas at high
temperatures and low pressure. This is
because they cannot be attracted to each other
if they are moving to fast and are very far apart
from each other.
 Volume
is directly proportional to the
moles of a gas at the same temperature
and pressure.

n = moles
 Remember
V1 = V2
n1
n2
1 mole of a gas at STP
occupies 22.4Liters
The Ideal Gas Law

Mathematical formula that relates the pressure, volume,
temperature and the number of particles(or moles) of a
gas.
PV = nRT

“R” is the gas constant,
R = 0.0821 atm.L
K.mole
The R value comes from using 1atm, 1 mole, 273K and
22.4 Liters as our comparison values.
Ideal gas law example
What is the pressure in atmospheres exerted by
a 0.500gram sample of nitrogen (N2) gas in 10.
Liter container at 25oC ?
*you must convert grams to moles
0.500g 1 mole
=
(and Celcius to Kelvin!!)
0.0178 moles
28.02 grams
PV = nRT
P (10L) = (0.0178moles)(0.0821 L.atm/K.mole) (298K)
P = 0.044 atm
You must use Liters, atm and kelvin in this
formula!!!!!
Other Derivations of the ideal gas law

To calculate the molar mass (M) of a gas
PV = nRT
Moles = n =
mass (m)
Molar Mass(M)
PV = m R T
M
Rearranging
Molar Mass = M = m R T
PV
Example:
A
6.39 gram sample of a gas occupies a
2.00L container at 117kPa and 35.1oC.
What is the molar mass of the gas?
Molar Mass = M = m R T
PV
M = (6.39g)(0.0821)(308.1)
(117/101.325)(2.00)
= 70.0 g/mole
Ideal Gas law Tutorial Online
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