CK-12-Physics - Intermediate

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 14. Thermodynamics


The Ideal Gas Law


The three gas laws we have previously discussed in addition to the relationship between volume, pressure, and mass,
lead to a more general statement of proportionality→PV∝mT. We can write this as an equation→PV=CmT,
but we can show that if the massmis used in the equation, then the constant of proportionalityCdepends upon the
particular gas. If however, the mass is expressed using the number of molesn, a constant of proportionality (known


as the universal gas constant)R= 8. (^314) molJ−K, is applicable for any gas.
Under these conditions theIdeal Gas Lawcan be expressed asPV=nRT, where the temperatureTmust be in
Kelvins. No gas exactly follows the Ideal Gas Law. But if the assumptions made for the kinetic theory of gases are
valid (that is, if the gas has high temperature and low pressure), then the Ideal Gas Law is applicable. Unless stated
otherwise, throughout this chapter we will assume that we deal with an ideal gas.
http://www.youtube.com/watch?v=PjLA0Fmi44A
Illustrative Example 14.1.1
What is the volume of one mole of any gas at one atmosphere of pressure at room temperature 20. 0 ◦C?
Answer:
Recall from earlier that 1atm= 1. 01 × (^105) mN 2 and that the temperature in the Ideal Gas Law equation must be in
Kelvin. Remember that
Tk=Tc+ 273 → 20 + 273 = 293 K
PV=nRT→


(


1. 01 × 105


N


m^2

)


V= ( 1. 00 mol)

(


8. 314


J


mol∗K

)


( 293 K)→


V=


( 1. 00 mol)

(



  1. (^314) molJ∗K)


(


293 K)


(



  1. 01 × (^105) mN 2


) = 0. 0241 = 24. 1 × 10 −^3


J∗m^2
N


24. 1 × 10 −^3


N∗m^3
N
→ 24. 1 × 10 −^3 m^3

What would the volume be at 0. 00 ◦C?


Answer: 22. 4 × 10 −^3 m^3


Illustrative Example 14.1.2


An ideal gas is contained in a rigid vessel at a pressure of 1. 15 atm at 29 ◦C. What is the pressure if the vessel is
heated to a temperature of 90◦C? Express the answer in atm.


Answer:


The volume and mass remain constant, thereforePV=nRT→TP=nRV =constant. Before calculating, don’t forget
to change the temperature to Kelvins!


Pi
Ti

=


Pf
Tf


1. 15 Patm
( 29. 0 + 273 )K

=


Pf
( 90. 0 + 273 )K
→Pf=

363 K


302 K


( 1. 15 Patm) = 1. 382 Patm→ 1. 38 Patm
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