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  4-40

Closed System Energy Analysis: Ideal Gases

4-58C  No, it isn't. This is because the first law relation Q - W = U  reduces to W  = 0 in this case since the system is

adiabatic (Q = 0) and U  = 0 for the isothermal processes of ideal gases. Therefore, this adiabatic system cannot receive

any net work at constant temperature.

4-59 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases.

 Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values

of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible, . 3 Constant specific

heats can be used for oxygen.

0 peke  

 Properties The specific heats of oxygen at the average temperature of (20+120)/2=70C=343 K are c p = 0.927 kJ/kgK and

cv = 0.667 kJ/kgK (Table A-2b).

 Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system.

The energy balance for a constant-volume process can be expressed as

)( 12in

energiesetc. potential, kinetic,internal,inChange

system

massandwork,heat, bynsfer energy tra Net

outin

T T mcU Q

 E  E  E 

v 

O2

T 1 = 20°C

T 2 = 120°C

Q

The energy balance during a constant-pressure process (such as in a piston-

cylinder device) can be expressed as

)( 12in

out,in

out,in

energiesetc. potential, kinetic,internal,inChange

system

massandwork,heat, bynsfer energy tra Net

outin

T T mc H Q

U W Q

U W Q

 E  E  E 

 p

b

b

O2

T 1 = 20°C

T 2 = 120°C

Q

since U  + W  b =  H  during a constant pressure quasi-equilibrium process.

Substituting for both cases,

kJ66.7

K )20K)(120kJ/kg667.0(kg)1()( 12constin, T T mcQv V  

 

kJ92.7K )20K)(120kJ/kg927.0(kg)1()( 12constin, T T mcQ  p P   

PROPRIETARY MATERIAL

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