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 = p∆T = nR∆T , 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 .