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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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