review2

1 1
2
2
W = 70.2 − 68.9 = −1.3( J )
• 3) two moles of a monatomic ideal gas are at contained at a pressure of 1 atm and temperature of 300K, 34.166j of heat are transferred to the gas as a result of which the gas expands and does 1216 joules of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.
a) constant pressure
∫ δW = ∫
a
b
b
a
PdV = P ∫ dV = P∆V = P0 × (Va − V0 )
a
b
= 101325 × (44.8 − 22.4) ×10−3 = 2.27(kJ )
nRT 1 × 8.314 × 273 V0 = = = 22.4(l ) p0 1 × 101325 Va = 2V0 = 44.8(l )
P = P0 = 1(atm) W = P∆V = nR∆T = 1 × 8.314(T − 273) = 832( J ) T = 373( K ) nRT 1 × 8.314 × 373 V = = = 30.6(l ) P 101325 ∆H = Q p = 3000( J ) ∆U = Q − W = 3000 − 832 = 2168( J ) ∵ ∆U = nC v (T − T0 ) = 1 × CV × 8.314(373 − 273) ∴ CV = 2.5 R
1) isothermally
P0 V0 101325 × 1× 10-3 n= = = 0.033( mol ) RT0 373 × 8.314 P1 = P0V0 = 0.5(atm) V1
V1 = V0 = 2(l )
nRT V1 ∫0 δW = ∫0 PdV = ∫0 V dV = nRT ln V0
a) Isothermal process
P0 = 1(atm), T0 = 298.15( K ) nRT0 1× 8.314 × 298.15 V0 = ( )= = 24.46(l ) P0 101325 Pa = 0.5(atm), Ta = 298.15( K ) P0V0 = PaV a V a = 48.92(l ) P0 W = nRT ln = 1× 8.314 × 298.15 ln 2 = 1.72(kJ ) Pa
The Thermodynamics of Materials and Its Applications
Review ቤተ መጻሕፍቲ ባይዱesson
For the former three chapters
Graduate Course, Nanchang University
Exercise for someone who wants to ……
1 1 1
= 101325 ×10 ln 2 = 70.2( J )
−3
P0V2γ = PV1γ 1 PV15 / 3 5 / 3 0.5 3 / 5 1 V2 = ( ) = ( ) × 2 = 1.32(l ) P0 1
∫
2
1
δW = ∫ PdV = P ∫ dV = 101325 ×(1.32 − 2) ×10 −3 = −68.9( J )
∫ δW = ∫
0
c
c
0
PdV = ∫ 6.643 ×10 − 4 V 2 + 0.6667dV = 2.214 × 10 − 4 [V 3 ]22..4 + 0.6667[V ]22..4 = 44 8 44 8
0
c
− 0.0149( J )
• 2)The initial state of a quantity of monatomic ideal gas is P=1atm V=1liter and T=373K. The gas is isothermally expanded to a volume of 2 liters and is then cooled at constant pressure to the volume V. this volume is such that a reversible adiabatic compression to a pressure of 1 atm returns the system its initial state. All of the changes of state are conducted revisable adiabatic. Calculate the total work done on or by the gas
b) isobaric process
Tb = 373.15( K ), Pb = 0.5(atm) nRT 1 × 8.314 × 373.15 Vb = ( )= = 61.24(l ) Pb 0.5 × 101325 W = P∆V = 101325 × 0.5 × (61.24 − 48.92) × 10 −3 = 0.624(kJ )
• the system then undergoes the following reversible cyclic process • e) an isobaric expansion to100℃, followed by • f) a decrease in pressure at constant volume to the pressure P atm, followed by • g) an isobaric compression at P atm 24.5 liters, followed by • h) an increase in pressure at constant volume to 1 atm • calculate the value of P which makes the work done on the gas during the first cycle equal to the work done by the gas during the second cycle.
b) constant volume
Pb = 2(atm),Vb = Va = 44.8(l )
∫ δW = ∫
a
Pc = 2(atm),Vc = 22.4(l )
b
b
a
PdV = 0
c) return to the initial state along the path p=6.643*10V2+0.6667
• 1) one mole of a monatomic ideal gas in the initial state T=273K, P=1atm, is subject to the following three processes, each of which is conducted reversibly. • a. a doubling of its volume at constant pressure • b.Then a doubling of its pressure at constant volume • c.Then a return to the initial state along the path p=6.643*10-4V2+0.6667 • Calculate the heat and work effects, which occur during each of the process.
isothermal
W = ∫ pdV = ∫
0 a a 0
P nRT dV = nRT ln 0 = 10 × 8.314 × 300 ln 10 = 57.43( kJ ) V Pa
Va =
nRT = 246(l ) Pa
Pa = 1(atm)
constant pressure
W = ∫ δW = ∫ PdV = Pa ∫ dV = P∆V = 1 × 101325 × (24.6 − 246) × 10 −3 = −22.43(kJ )
a a a b b b
Vb = 24.6(l ) Pb = 10(atm)
constant volume
W = 0 V b = 24 . 6 ( l ) Pb = 10 ( atm )
The total W
W = 57.43 − 22.43 = 35( kJ )
• 6) one mole of an ideal gas at 25℃ and 1atm undergoes the following reversible conducted cycle: • a)an isothermal expansion to 0.5 atm, followed by • b)an isobaric expansion to 100℃ followed by • c)an isothermal compression to 1 atm followed by • d)an isobaric compression to 25℃
C) Isothermal process
Pc = 1(atm), Tc = 373.15( K ) nRTc 1 × 8.314 × 373.15 Vc = ( )= = 30.62(l ) Pc 101325 Pb W = nRT ln = 1 × 8.314 × 373.15 ln 0.5 = −2.15(kJ ) Pc