P20 082

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82.

(a) From Table 20-3, C

V

=

5
2

R and C

p

=

7
2

R. Thus, Eq. 20-46 yields

Q = nC

p

T = (3.00)

7

2

(8.31)

(40.0) = 3490 J .

(b) Eq. 20-45 leads to

E

int

= nC

V

T = (3.00)

5

2

(8.31)

(40.0) = 2493

2490 J .

(c) From either W = Q

E

int

or W = pT = nRT , we find W = 997J.

(d) Eq. 20-24 is written in more convenient form (for this problem) in Eq. 20-38. Thus, we obtain

K

trans

= ∆ (N K

avg

) = n

3

2

R

T

1500 J .


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