Starting with Equation 18.11, show that the total pressure P in a container filled with a mixture of several ideal gases is P = P 1 + P 2 + P 3 + ⋯ , where P 1 , P 2 , ⋯ are the pressures that each gas would exert if it alone filled the container. (These individual pressures are called the partial pressures of the respective gases.) This result is known as Dalton’s law of partial pressures .
Starting with Equation 18.11, show that the total pressure P in a container filled with a mixture of several ideal gases is P = P 1 + P 2 + P 3 + ⋯ , where P 1 , P 2 , ⋯ are the pressures that each gas would exert if it alone filled the container. (These individual pressures are called the partial pressures of the respective gases.) This result is known as Dalton’s law of partial pressures .
Solution Summary: The author illustrates how the total pressure in a container filled with several ideal gases is given by left.
Starting with Equation 18.11, show that the total pressure P in a container filled with a mixture of several ideal gases is
P
=
P
1
+
P
2
+
P
3
+
⋯
, where
P
1
,
P
2
,
⋯
are the pressures that each gas would exert if it alone filled the container. (These individual pressures are called the partial pressures of the respective gases.) This result is known as Dalton’s law of partial pressures.
1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm.
Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from
the center of the sphere.
(a) =
=
(b) E =
(c)Ẻ =
=
NC NC NC
1.
A long silver rod of radius 3.5 cm has a charge of -3.9
ис
on its surface. Here ŕ is a unit vector
ст
directed perpendicularly away from the axis of the rod as shown in the figure.
(a) Find the electric field at a point 5 cm from the center of the rod (an outside point).
E =
N
C
(b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point)
E=0
Think & Prepare
N
C
1. Is there a symmetry in the charge distribution? What kind of symmetry?
2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ
from a?
1. Determine the electric flux through each surface whose cross-section is shown below.
55
S₂
-29
S5
SA
S3
+ 9
Enter your answer in terms of q and ε
Φ
(a) s₁
(b) s₂
=
-29
(C) Φ
զ
Ερ
(d) SA
=
(e) $5
(f) Sa
$6
=
II
✓
-29
S6
+39
Chapter 19 Solutions
Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
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