1. No category

Sample problems chapter 21
20.
A 1.00-mol sample of a diatomic ideal gas has pressure P and volume V.
When the gas is heated, its pressure triples and its volume doubles. This heating
process includes two steps, the first at constant pressure and the second at
constant volume. Determine the amount of energy transferred to the gas by heat.
P21.20 Q = ( nC P ∆T )isobaric + ( nCV ∆T )isovolum etric
In the isobaric process, V doubles so T must double, to 2Ti.
In the isovolumetric process, P triples so T changes from 2Ti to 6Ti.
7 
5 
Q = n  R  ( 2Ti − Ti) + n  R  ( 6Ti − 2Ti) = 13.5nRTi = 13.5PV
2 
2 
25.
A 2.00-mol sample of a diatomic ideal gas expands slowly and
adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume
of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and
final temperatures? (c) Find Q, W, and ∆Eint.
P21.25
(a)
γ
γ
PV
i i = PfV f
γ
1.40
V 
 12.0 
= 1.39 atm
Pf = Pi i  = 5.00 atm 
 30.0
 Vf
(b)
(
Tf =
(c)
)(
5.00 1.013 × 105 Pa 12.0 × 10−3 m
PV
i i
=
Ti =
nR
2.00 m ol( 8.314 J m ol⋅ K )
PfV f
nR
=
(
)(
3
1.39 1.013 × 105 Pa 30.0 × 10−3 m
2.00 m ol( 8.314 J m ol⋅ K )
)=
3
)=
365 K
253 K
The process is adiabatic: Q = 0
γ = 1.40 =
C P R + CV
5
=
, CV = R
CV
CV
2
5

∆Eint = nC V ∆T = 2.00 m ol ( 8.314 J m ol⋅ K ) ( 253 K − 365 K ) = −4.66 kJ
2

W = ∆Eint − Q = −4.66 kJ− 0 = −4.66 kJ
51.
The function Eint = 3.50nRT describes the internal energy of a certain ideal
gas. A sample comprising 2.00 mol of the gas always starts at pressure 100 kPa
and temperature 300 K. For each one of the following processes, determine the
final pressure, volume, and temperature; the change in internal energy of the gas;
the energy added to the gas by heat; and the work done on the gas. (a) The gas is
heated at constant pressure to 400 K. (b) The gas is heated at constant volume to
400 K. (c) The gas is compressed at constant temperature to 120 kPa. (d) The gas
is compressed adiabatically to 120 kPa.
P21.51
(a)
Pf = 100 kPa
Vf =
nRT f
Pf
=
T f = 400 K
2.00 m ol( 8.314 J m ol⋅ K ) ( 400 K )
100 × 103 Pa
= 0.066 5 m
3
= 66.5 L
∆Eint = ( 3.50) nR ∆T = 3.50( 2.00 m ol) ( 8.314 J m ol⋅ K ) ( 100 K ) = 5.82 kJ
W = − P∆V = − nR ∆T = − ( 2.00 m ol) ( 8.314 J m ol⋅ K ) ( 100 K ) = −1.66 kJ
Q = ∆Eint − W = 5.82 kJ+ 1.66 kJ= 7.48 kJ
(b)
T f = 400 K
V f = Vi =
nRTi 2.00 m ol( 8.314 J m ol⋅ K ) ( 300 K )
=
= 0.049 9 m
Pi
100 × 103 Pa
 Tf 
 400 K 
Pf = Pi  = 100 kPa
= 133 kPa
 300 K 
T
 i
V = constant
3
= 49.9 L
W = − ∫ PdV = 0 since
∆Eint = 5.82 kJ as in part (a)
Q = ∆Eint − W = 5.82 kJ− 0 = 5.82 kJ
(c)
Pf = 120 kPa
T f = 300 K
P 
 100 kPa
V f = V i i  = 49.9 L 
= 41.6 L
 120 kPa
P
 f
∆Eint = ( 3.50) nR ∆T = 0 since
T = constant
Vf
W = − ∫ PdV = − nRTi ∫
Vi
P 
 V f
dV
= − nRTiln   = − nRTiln  i 
V
 Vi 
 Pf 
 100 kPa
W = − ( 2.00 m ol) ( 8.314 J m ol⋅ K ) ( 300 K ) ln 
= +909 J
 120 kPa
Q = ∆Eint − W = 0 − 910 J= −909 J
(d)
γ =
Pf = 120 kPa
γ
C P CV + R 3.50R + R 4.50 9
=
=
=
=
CV
CV
3.50R
3.50 7
P 
V f = V i i 
 Pf 
γ
PfV f = PiV i : so
1γ
 100 kPa
= 49.9 L 
 120 kPa
79
= 43.3 L
 PfV f 
 120 kPa  43.3 L 
= 300 K 
= 312 K
T f = Ti

 100 kPa  49.9 L 
 PV
i i
∆Eint = ( 3.50) nR ∆T = 3.50( 2.00 m ol) ( 8.314 J m ol⋅ K ) ( 12.4 K ) = 722 J
Q = 0
( adiabatic process)
W = −Q + ∆Eint = 0 + 722 J= +722 J
Sample Problems Chapter 22
3.
A particular heat engine has a useful power output of 5.00 kW and an
efficiency of 25.0%. The engine expels 8 000 J of exhaust energy in each cycle.
Find (a) the energy taken in during each cycle and (b) the time interval for each
cycle.
P22.3
(a)
We have e =
W eng
Qh
=
Qh − Qc
Qh
= 1−
Qc
Qh
= 0.250
with Q c = 8 000 J, we have Q h = 10.7 kJ
(b)
W eng = Q h − Q c = 2 667 J
and from P =
W eng
∆t
, we have ∆t=
W eng
P
=
2 667 J
= 0.533 s .
5 000 J s
13.
An ideal gas is taken through a Carnot cycle. The isothermal expansion
occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas
takes in 1 200 J of energy from the hot reservoir during the isothermal expansion.
Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net
work done by the gas in each cycle.
P22.13 Isothermal expansion at
Th = 523 K
Isothermal compression at
Tc = 323 K
Gas absorbs 1 200 J during expansion.
(a)
T 
 323
Q c = Q h  c  = 1200 J
= 741 J
 523
 Th 
(b)
W eng = Q h − Q c = (1200 − 741) J= 459 J