Homework, Chapter 14, Thermodynamics

Ch 14: 1, 17, 26, 27, 37, 44, 46, 52, 58

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**14.1** An ideal gas is sealed in a rigid container at
25°C and 1.0 atm. What will its temperature be when the pressure
is incresed to 2.0 atm?

What will its pressure be when the temperature is increased to
50°C

The ideal gas law isPV = nRT Being in a rigid container, the gas’ volume does

notchange; V = V_{o}= constant. That means we can use the ideal gas law asT/P = T _{o}/P_{o}or

T = P [ T _{o}/P_{o}] = [ P/P_{o}] T_{o}Remember, these temperatures must be

absolutetemperatures,T _{o}= 25°C = 302 K

T = [ P/P_{o}] T_{o}= [^{2.0 atm}/_{1.0 atm}] [302 K] = [ 2 ] [ 302 K] = 604 K

T = 604 K = (604 - 273)°C =331°C = TOr, while V = const, we can write the ideal gas law as

P/T = P _{o}/T_{o}or

P = T [ P _{o}/T_{o }] = [ T/T_{o}] P_{o}Remember, these temperatures must be

absolutetemperatures, measured in Kelvins, K.T _{o}= 25°C = 302 K and T = 50°C = 327 K_{ }P = [^{T}/_{T}_{o }] P_{o}

P = [^{ }^{327 K}/_{302 K}] (1 atm)

P = 1.08 atm

**14.17** Gas at 1.5 atm expands from 2.5 liters to 3.5 liters.
How much work is done by the gas?

Work is the area under the curve on a p-V diagram; for constant pressure, this isW = p V

W = (1.5 atm)(1.0 l)

W = 1.5 atm-l = 1.5 l-atmWhile this is certainly true, a conversion to units of joules is called for,

1 l-atm = ( 10 ^{- 3}m^{3 }) (1.013 x 10^{5}^{N}/_{m}2) = 1.013 x 10^{2}N-m = 101.3 J

W = 1.5 l-atm [^{101.3 J}/_{1 l-atm}] =152 J = W

W =152 J

**14.26** Under a constant pressure of 1.5 atm, a gas expands
from 2.0 liters to 3.0 liters while 500 J of energy flows into
it.

(a) How much work is done by the gas?

(b) What is the change in its internal energy?

Part (a) of this question isto question 14.17. We already know the answer to that; the work done by the gas isidenticalW = 152 J

As the gas does 152 J of work, it absorbs or receives 500 J of energy. That means its net increase in internal energy isE

_{int}= Q - W = 500 J - 152 J = 348 J

E_{int}= 348 J

(a) How much work is doneonthe gas?

(b) What is the change in its internal energy?Work is the area under the curve on a p-V diagram; for constant pressure, this is

W = p V

W = (2.25 atm) (- 0.50 l)

W = - 1.125 l-atm

W = - 1.125 l-atm [^{101.3 J}/_{1 l-atm}] = - 114 JAs the gas does W = - 114 J of work (which is equivalent to having 114 J of work done

onit), it also gives up 375 J of energy (which means Q = - 375 J). That means its net change in internal energy isE _{int}= Q - W = - 375 J - (- 114 J)

E_{int}= - 261 J

**14.37** After 1.0 mol of an ideal gas initially at 0°C
and 1.0 atm expands __isothermally__ to twice its initial volume,
it is compressed __isobarically__ back to its original volume.
What is the net work done by or on the gas? What is its final
temperature?

This calls for a p-V diagram.

When the gas expands

isothermallyfrom state1to state2, the work done by the gas is given by Equation 14.17,W _{isotherm}= n R T ln [V_{f}/V_{i}]or

W _{isotherm}= n R T ln [V_{2}/V_{1}]From the problem statement, we know

V _{2}= 2 V_{1}From this, we can calculate the work done by the gas,

W _{isotherm}= (1.0 mol)(8.314^{J}/_{mol-K}) (273 K) ln[2^{ }V_{1 }/V_{1}]

W_{isotherm}= (1.0 mol)(8.314^{J}/_{mol-K}(273 K) ln[ 2 ]

W_{isotherm}= (1.0 mol)(8.314^{J}/_{mol-K }) (273 K) 0.693

W_{isotherm}= 1,573 JThis is the area under the curve from state

1to state2As the gas is compressed from state

2to state3the work done is, as always, the area under the curve.

W_{isobar}= p V = p (V_{o}- 2 V_{o}) = - p V_{o}We must find p, the pressure for state

2or state3. Since we got to state2by anisothermalexpansion from state1, we know the temperature at state2must be the same as at state1; T_{1}= T_{2}= 273 K. We can then apply the ideal gas law or, more simply, just usepV = constant ( for constant T )

p_{2}V_{2}= p_{1}V_{1}

p_{2}= p_{1}(V_{1}/V_{2}) = (1.0 atm) (V_{1}/2 V_{1}) = 0.5 atm = p_{2}

p_{2}= p_{1}(^{1}/_{2}) = 0.5 atmTo find the work done, W

_{isobar}= - pV_{o}, we need a numerical value for V_{o}, the initial volume. We could also call this V_{1}, the volume at state1. To find this initial volume V_{o}or V_{1}, we use the ideal gas law,pV = nRT Again, remember that temperatures are often given in degrees Celsius because that is “convenient” but temperatures in the ideal gas law must be

absolutetemperatures or temperatures on the Kelvin scale!pV = nRT

(1.0 atm) (V_{o}) = (1.0 mol) (8.314^{J}/_{mol-K}) (273 K)

(1.0 atm) [ (1.013 x 10^{5}^{N}/_{m}2)/atm ] (V_{o}) = (1.0 mol) (8.314^{J}/_{mol-K }) (273 K)

[ 1.013 x 10^{5}^{N}/_{m}2 ] (V_{o}) = (1.0 mol)(8.314^{J}/_{mol-K }) (273 K)

V_{o}= 2.24 x 10^{- 2}m^{3}

V_{1}= V_{o}= 2.24 x 10^{- 2}m^{3}Now we can go back to calculating the work done on/by the gas,

W _{isobar}= - p V_{o}= - (0.5 atm)(2.24 x 10^{- 2}m^{3})

W_{isobar}= - p V_{o}= - (0.5 x 1.013 x 10^{5}^{N}/_{m}2 )(2.24 x 10^{- 2}m^{3})

W_{isobar}= - 1,135 J

Now we can add these two amounts of work together for thework done.netW

_{net}= W_{12}+ W_{23}

W_{net}= W_{isotherm}+ W_{isobar}

W_{net}= 1,573 J + (-1,135 J)

W_{net}= 438 J

This

work is the area shown on the graph above.net

**14.44** A heat engine absorbs 6 J from a source of heat and
does 2 J of mechanical work. What is its efficiency?

Eff = 2 J / 6 J = 0.33

Eff = 0.33

**14.46** A reversible heat engine whose efficience is 15
percent does 180 J of work. How much heat does it absorb from the hot
reservoir?

Q_{h} = W / Eff

Q_{h} = ^{180
J}/_{0.15}

Q_{h} = 1,200 J

Remember, all these temperatures must betemperatures. That means for our initial Carnot engine,absoluteT _{h}= 650°C = (650 + 273) K = 923 KFirst, find T

_{c}the temperature of the cold reservoir. That will remain the same.Eff = 1 - T _{c}/T_{h}= 0.30

T_{c}/T_{h}= 1 - 0.30 = 0.70

T_{c}= 0.70 T_{h}= (0.70) (923 K) = 646 K

( T_{c}= 373°C )Now, with this as the cold reservoir (T

_{c}= 646 K), what temperature for the hot reservoir (T_{h}= ?) will give an efficiency of 0.35?Eff = 1 - T _{c}/T_{h}= 0.35

T_{c}/T_{h}= 1 - 0.35 = 0.65

T_{h}= T_{c}/0.65T

_{h}=^{646 K}/_{0.65}

T_{h}= 993 K

( T_{h}= 620°C )

**
**

Tothe operation of a real engine, like a gasoline engine, we may treat it as if it were a reversible or Carnot engine. For such a Carnot engine, the efficiency isestimateEff = 1 - T _{c}/T_{h}The exhaust gases leave at the cold temperature of T_{c}= 125°C = 398 KT _{c}/T_{h}= 0.55

T_{c}/T_{h}= 1.0 - 0.55 = 0.45

T_{h}=^{398 K}/_{0.45}

T_{h}= 884 K

( T_{h}= 611°C )

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(c) 2000, Doug Davis; all rights reserved