The piston/cylinder arrangement shown in Fig. Q3 contains air at a pressure of 3
ID: 1939923 • Letter: T
Question
The piston/cylinder arrangement shown in Fig. Q3 contains air at a pressure of 300 kPa and a temperature of 500 K (this is state 1). The mass of the air is 3 kg. The air is expanded in a constant pressure process to twice its original volume (this is state 2). The piston is then locked in place with a pin, and heat is transferred until it reaches a final temperature of 500 K (this is state 3).
Calculate:
(a) the pressure at state 3 (p3);
(b) the temperature at state 2 (T2);
(c) the work transfer between states 1 and 2 (W)
1-2
(d) the work transfer between states 2 and 3 (W)
2-3
(e) the heat transfer between states 1 and 2 (Q)
1-2
(f) the heat transfer between states 2 and 3 (Q)
2-3
The properties of air may be related to one another by the equation
pV = mRT
where, p = pressure; V =volume; m =mass; R =specific gas constant; and T =
temperature.
Explanation / Answer
state 1 :
300 kpa
500 k
m=3kg
from state 1 we can have :
PV= mRT
V = mRT/P
Volume of state 1 : 1.435
state 2 :
300 kpa
V2 = 2 V1 = 2.87
3kg
state 3 :
v3 = v2 = 2.87 ( contant volume after the pin was attached )
3 kg
500 K
from state 3 we can have :
PV = mRT
a ) P = mrT/V = 150 KPa
b ) what we know in state two is :
state 2 :
300 kpa
V2 = 2 V1 = 2.87
3kg
PV= mRT
T = PV/mR = 1000 K
c ) boundary work at a constant pressure = P ( V2 - V1 ) = 300 ( 2.87 - 1.435 ) = 430.5 KJ
d ) since this is a constant volume process, the work done is zero.
e ) first law of thermodynamics :
Q-W = U + KE + PE
Q = W + U ( changes in potential and kinetic energies is almost zero since the system is not moving )
Q = W + mCv(T2 - T1 ) ( because U can be approximated to Cv(T2 - T1 ) - Cv from tables is 0.717 )
Q = 430.5 + 1075.5 =1506 KJ
f )first law of thermodynamics :
Q-W = U + KE + PE
Q = U ( changes in potential and kinetic energies is almost zero since the system is not moving )
Q = 1075.5 KJ
Good luck ,
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