CK-12-Physics-Concepts - Intermediate

http://www.ck12.org Chapter 11. The Fluid States

TABLE11.1:(continued)

atm milliliters moles Kelvin 82.1 mL•atm/mol•K

Since the product of (liters)(atm) can be converted to joules, we also have a value forRwhere liters×atm have
been converted to joules,R= 8 .314 J/mol·K. The two common values of theuniversal gas lawRconstant
are 0.0821 L·atm/mol·K andR= 8 .314 J/mol·K.

Most universal gas law problems are calculated at STP. STP stands forstandard temperature and pressure,which is
the most commonly calculated temperature and pressure value. STP is defined as 1.00 atm and 0°C, or 273 K.

Example Problem:Determine the volume of 1.00 mol of any gas at STP.

Solution:First isolate V from PV=nRT. Then plug in known values and solve.

V=

nRT

P

=

( 1 .00 mol)( 0 .0821 L·atm/mol·K)(273 K)

( 1 .00 atm)

= 22 .4 liters

For any gas at STP, one mole has a volume of 22.4 liters. This can be an extremely convenient conversion factor.

Example Problem:A sample of oxygen gas occupies 10.0 liters at STP. How many moles of oxygen are in the
container?

Solution:

n=

PV

RT

=

( 1 .00 atm)( 10 .0 L)

( 0 .0821 L·atm/mol·K)(273 K)

= 0 .446 moles

Summary

• The universal gas law isPV=nRT, wherePis pressure,Vis volume,nis number of moles,Ris the universal
  gas law constant, andTis the absolute temperature.


• The value ofRvaries depending on the units used forPandV. Two common values are 0.0821 L·atm/mol·K
  andR= 8 .314 J/mol·K.


• STP is standard temperature and pressure; 273 K and 1.00 atm.


• One mole of a gas at STP has a volume of 22.4 liters.

Practice

Questions

The following video discusses the constantR. Use this resource to answer the questions that follow.

http://youtu.be/kBgzzwCTddc

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/64121