41.

(a) From ρ

H

= 0.35ρ = n

p

m

p

, we get the proton number density n

p

:

n

p

=

0.35ρ

m

p

=

(0.35)(1.5

× 10

5

kg/m

3

)

1.67

× 10

−27

kg

= 3.14

× 10

31

m

−3

.

(b) From Chapter 20 (see Eq. 20-9), we have

N

V

=

p

kT

=

1.01

× 10

5

Pa

(1.38

× 10

−23

J/K)(273 K)

= 2.68

× 10

25

m

−3

for an ideal gas under “standard conditions.” Thus,

n

p

(N/V )

=

3.14

× 10

31

m

−3

2.44

× 10

25

m

−3

= 1.2

× 10

6

.