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Physics 152 Course Notes for Mon. 9/20, 1999 Readings: H,R&W: Ch 20, Secs. 9-11, Serway: Ch 21, Secs. 4,6. Homework: 1) One mole of an ideal monatomic gas initially at 300 K and one atmosphere of pressure is compressed quasi-statically and adiabatically to one-fourth its initial volume. Find its final temperature and pressure. 2) In a constant-volume process, 209J of heat is transferred to one mole of an ideal, monatomic gas initially at 300 K. Find: a) The work done by the gas, b) the increase in internal energy of the gas, and c) the final temperature of the gas. 3) Show how to get the values in the last table at the end of the notes on the back of this page.

______________________________________________________________________________

Adiabatic Process: An adiabatic process is one where no heat is transferred between the system and its surroundings. For an adiabatic process for an ideal gas, we can write:

PV = constant , or equivalently, TV(

-1)

= constant ,

where the constant, , is the ratio: = C p CV and is equal to 1.67 for a monatomic ideal gases. Summary of Relationships and Processes: Consider the figure at the right. Four processes are contained in these five paths: Isobaric: Paths 1 and 3.

P 1

p = constant W = pV = p( V f - V i ) Q = nC p T = nC p ( T f - T i )

First Law: E = Q - W = nC p T - pV . Isothermal: Path 2

4 2'

2

3 V

T = constant ,

E = nCV T =

3 nRT = 0 , 2

W = nRT ln

Vf Vi

,

First Law: E = 0 Adiabatic: Path 2'

Q=W.

Q = 0,

PV = constant ,

First Law: E = -W

Isovolumetric: Path 4

V = constant ,

W = 0,

Q = nCV T , First Law: E = Q

There are two cyclic processes: Path 1234, or path 12'34. For cyclic processes, the initial and final temperatures are the same, and E = 0 . The First Law for cyclic processes becomes: Q = W . For all processes for an ideal gas, we can use the equation of state: PV = nRT

Example: Three moles of Argon gas, initially at 20oC, occupies a volume of 10L. The gas undergoes a slow expansion at constant pressure to a volume of 25L. Then the gas expands adiabatically until it returns to its initial temperature. a) Show the process on a P-V diagram b) Determine the macroscopic variables V, p, and T for the beginning and end points of each process . c) Determine E, Q, and W for the two processes in your P-V diagram. Solution: a) The P-V diagram is shown at the right.

P AB: Isobaric B b) The table shows what we are given in the problem for the beginning and end points of each process. To find the pressure at PA=PC point A, apply the perfect gas law: PV=nRT. This will result in PA=7.31x105N/m2. This is also the p V T pressure at point B since process AB is BC: adiabati c isobaric. Knowing the pressure and A 10 -2 m 3 293K volume for point B, we apply the perfect gas law again to obtain the temperature at PC C point B. This results in: TB=732.5K. V 2.5x This only leaves the pressure and volume B VB VA VC at point C to be determined. To find the 10 -2 m 3 volume at point C, we know that for adiabatic processes, the equation,

A

C

293K

TV(

-1)

following equality:

= constant , is valid. So, we can write for points B and C the

( -1 ( -1 T BV B ) = T C V C ) .

Solving for the volume, VC, gives:

T -1 732.5K 1.67-1 VC = B VB = 2.5 × 10 -2 m3 = 9.8 × 10 -2 m3 . 293K TC

1

1

(

)

Since you now know both T and V at point C, you can now apply the ideal gas law to get PC=7.45x104N/m2. The final table looks as follows: To get more practice, try to see if you can obtain the values in the table below for the internal energy, heat, and work for the three processes, AB, BC, and AC (this is homework problem #3 above):

p

7.31x

V

T

293K

A 10 5 N/m 2 10 -2 m3 B 7.31x

2.5x

E

Q

2.74x 10 4 J

W

1.10x 10 4 J 1.65x 10 4 J 2.75x 10 4 J

-2 3 10 5 N/m 2 10 m

732.5K

AB

1.64x 10 4 J -1.64x 10 4 J

C 7.45x

9.8x

-2 3 10 4 N/m 2 10 m

293K

BC

0

AC

2.74x 0 10 4 J

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