10.3 Solutions
433
10.17
ρ 1 T 1
P 1
=
ρ 2 T 2
P 2
M 1 = Vρ 1 ,
M 2 = V ρ 2
∴
M 1
M 2
=
ρ 1
ρ 2
=
P 1 T 2
P 2 T 1
∴ M 2 =
P 2
P 1
·
T 1
T 2
M 1 =
50
76
×
273
263
× 175
= 119.5 kg
10.18 Let n moles be total mass of air in the two bulbs.
Initially, T = 273 + 20 = 293 K, P = 76 cm of Hg, V = V 1 + V 2 =
100 + 500 = 600 cc.
n =
PV
RT
=
70 × 600
293R
Finally, let n 1 and n 2 moles be the mass of air in the small and large bulb,
respectively. Under new conditions
n 1 =
P 1 V 1
RT 1
=
100P 1
293R
n 2 =
P 2 V 2
RT 2
=
500P 2
293R
(∵ P 2 = P 1 )
But n = n 1 + n 2 =
100P 1
293R
+
500P 1
373R
=
70 × 600
293R
Cancelling off R, we find P 1 = 85.23 cm of Hg.
10.3.3 Heat Transfer
10.19 (a) In the first slab, heat flow is given by
−
dQ 1
dt
=
k 1 A (T 1 − T )
d 1
(1)
In the second slab, heat flow is given by
−
dQ 2
dt
=
k 2 A(T − T 2 )
d 2
(2)
Now the continuity of heat flow requires that heat flow must be the same
in both the slabs (Fig. 10.7). Thus
433
10.17
ρ 1 T 1
P 1
=
ρ 2 T 2
P 2
M 1 = Vρ 1 ,
M 2 = V ρ 2
∴
M 1
M 2
=
ρ 1
ρ 2
=
P 1 T 2
P 2 T 1
∴ M 2 =
P 2
P 1
·
T 1
T 2
M 1 =
50
76
×
273
263
× 175
= 119.5 kg
10.18 Let n moles be total mass of air in the two bulbs.
Initially, T = 273 + 20 = 293 K, P = 76 cm of Hg, V = V 1 + V 2 =
100 + 500 = 600 cc.
n =
PV
RT
=
70 × 600
293R
Finally, let n 1 and n 2 moles be the mass of air in the small and large bulb,
respectively. Under new conditions
n 1 =
P 1 V 1
RT 1
=
100P 1
293R
n 2 =
P 2 V 2
RT 2
=
500P 2
293R
(∵ P 2 = P 1 )
But n = n 1 + n 2 =
100P 1
293R
+
500P 1
373R
=
70 × 600
293R
Cancelling off R, we find P 1 = 85.23 cm of Hg.
10.3.3 Heat Transfer
10.19 (a) In the first slab, heat flow is given by
−
dQ 1
dt
=
k 1 A (T 1 − T )
d 1
(1)
In the second slab, heat flow is given by
−
dQ 2
dt
=
k 2 A(T − T 2 )
d 2
(2)
Now the continuity of heat flow requires that heat flow must be the same
in both the slabs (Fig. 10.7). Thus