题目
又 T = 40 eC_ RT Up~ V ^b~ T^5V(V +b)8.3.4x313.15 3.2217~ 1.6× 10^4 - 2.985 × 10^5 ^ 3.3.1505 × 1.6× 10^4 × (1.6× 10^4 + 2.985 × 10^5) =1.401 × IO7 Pa = 133 atm > p ⅛ = IOO atm故储气罐会发生爆炸。2.16.乙烷是重要的化工原料,也可以作为冷冻剂。现装满290 K、2.4. MPa乙烷蒸气的钢瓶,不小 心接近火源被加热至478 K,而钢瓶的安全工作压力为4.5 MPa,问钢瓶是否会发生爆炸?(用RK 方程计算)解:由附录査得乙烷的临界参数。Tc=30.br>5.4., Pc=4.884MPa, Vc=1.48× 104m3∕mol; g>=0.091) T=290K, P=2∙48Mpa∙/ Tr=T∕Tt∙=0.95 Pr=P∕Pv=0.51・•・使用普遍化第二维里系数法。Bo-0.083-0.422/0.9516 --0.375B#-0.139-0.172/0.9542 --0.074«/>—■ B° + flrfΓ ■ -0.375 + 0.098x(-0.074) ■ -0.3823/. ½ = :R = 16..74 cm3 - moΓ1 Piz2=z? + ωi1 = 0.675 + 0.225 × 0.2 = 0.720.-.V2 = 149.9 cn√.mol-故 M∕ = V<z-%=149∙9-166∙74 = -16∙84 cm3 mol l过程①用普遍化压缩因子法查图得 也工.=_2 34 I = -O 89RTC ■ "^7?.仝血+ q<1.2∙54RA R7; R7;・•. H, = -2.54. 8.314 × 304.2 = -6424 kJ - kmoΓ查图得:IAL-L = -140R^(Sr):(Srl_R R RSf= -1.58.χ 8.3.4 = -13.14kJ kmoΓ, - K1过程②C: =2.br>6.75 + 4..258 XIO-ST+ (-14.25×1 O^)T2::;Cw279.6 KJ kmol-ASJ = f∙w∙,sSLJT-RIn= 3.635 kJ kmoΓ, K'1 P JWiS T O53过程③用普压法:= + ω^- = -1.72-0.225 X 0.20 = -1.78R7; R7; R7;.= !⅛1.+ O LlLL = -0.95 - 0.225 × 0.20 = -1.02R R R/. λ∕f =-1.78 × 314.X 304.2 = -4501.8 kJ kmolSf=-1.02×8.3I4 = -8.48kJ kmol l ∙K1ΔH = (-H:) + (∕∕f) + ∆Hld= 6424 + 3279.6-4501.8 = 5202 kJ kmol'ΔS = (-Sf) +∆Sp,d +(Sf) = I 3.4 + 3.635 - 8.48 = 83 kJ - kmoΓ1 ∙ KjΔt∕ = ΔH-Δ∕7V = 5.02-115.79.x149.9一8.053×219.3 = 4601 kJ kmoll⏺∆∕y 2.17(ln P-I)X 4.184Tap (0.930-7;,,)・..Tll = 2.3.15 ÷ (-0.5) = 272.65 K由表中数据知,正丁烷的临界压力,临界温度,如下:P= 3.786MPa = 38atmΓ =425.13K⏺2.17.ln 37.86-1)0.93-2.2.65425.13× 4.184 =5.72360.2888J - mol⏺H=I9.8205 kJ moΓ(2) T = 13.br>7.78 + 273.15 = 410.93K 蒸发热可按WatSon经验式讣算:I-T °”P)士竺= 0.6413425.13=出生="425.13∏ =19.8.05X(^)^ =8.0415 kJ5「(3)由 ClaPCyrOn 方程知,d严 _ △/TF T∆vCiT由数据拟合In 〃哪〜丄得到In/^ =21.752-28.8.4/Γ
又 T = 40 eC_ RT Up~ V ^b~ T^5V(V +b)
8.
3.4x313.15 3.2217~
1.6× 10^4 -
2.985 × 10^5 ^
3.3.1505 × 1.6× 10^4 × (1.6× 10^4 + 2.985 × 10^5) =1.401 × IO7 Pa = 133 atm > p ⅛ = IOO atm故储气罐会发生爆炸。
2.1
6.乙烷是重要的化工原料,也可以作为冷冻剂。现装满290 K、2.
4. MPa乙烷蒸气的钢瓶,不小 心接近火源被加热至478 K,而钢瓶的安全工作压力为4.5 MPa,问钢瓶是否会发生爆炸?(用RK 方程计算)解:由附录査得乙烷的临界参数。Tc=3
0.br>5.
4., Pc=4.884MPa, Vc=
1.48× 104m3∕mol; g>=0.091) T=290K, P=2∙48Mpa∙/ Tr=T∕Tt∙=
0.95 Pr=P∕Pv=0.51・•・使用普遍化第二维里系数法。Bo-
0.083-0.422/0.9516 --0.375B#-
0.139-0.172/0.9542 --0.074«/>—■ B° + flrfΓ ■ -
0.375 + 0.098x(-0.074) ■ -0.3823/. ½ = :R = 1
6..74 cm3 - moΓ1 Piz2=z? + ωi1 =
0.675 + 0.225 × 0.2 = 0.720.-.V2 = 14
9.9 cn√.mol-故 M∕ = V<z-%=149∙9-166∙74 = -16∙84 cm3 mol l过程①用普遍化压缩因子法查图得 也工.=_2 34 I = -O 89RTC ■ "^7?.仝血+ q<
1.2∙54RA R7; R7;・•. H, = -
2.5
4.
8.314 × 304.2 = -6424 kJ - kmoΓ查图得:IAL-L = -140R^(Sr):(Srl_R R RSf= -
1.5
8.χ 8.
3.4 = -13.14kJ kmoΓ, - K1过程②C: =
2.br>6.75 +
4..258 XIO-ST+ (-14.25×1 O^)T2::;Cw27
9.6 KJ kmol-ASJ = f∙w∙,sSLJT-RIn=
3.635 kJ kmoΓ, K'1 P JWiS T O53过程③用普压法:= + ω^- = -
1.72-
0.225 X 0.20 = -1.78R7; R7; R7;.= !⅛
1.+ O LlLL = -
0.95 - 0.225 × 0.20 = -1.02R R R/. λ∕f =-
1.78 × 31
4.X 304.2 = -4501.8 kJ kmolSf=-
1.02×
8.3I4 = -8.48kJ kmol l ∙K1ΔH = (-H:) + (∕∕f) + ∆Hld= 6424 + 327
9.6-450
1.8 = 5202 kJ kmol'ΔS = (-Sf) +∆Sp,d +(Sf) = I
3.4 + 3.635 -
8.48 = 83 kJ - kmoΓ1 ∙ KjΔt∕ = ΔH-Δ∕7V =
5.02-115.7
9.x149.9一
8.053×219.3 = 4601 kJ kmoll⏺∆∕y
2.17(ln P-I)X
4.184Tap (
0.930-7;,,)・..Tll =
2.
3.15 ÷ (-
0.5) = 272.65 K由表中数据知,正丁烷的临界压力,临界温度,如下:P=
3.786MPa = 38atmΓ =42
5.13K⏺
2.1
7.ln 37.86-1)
0.93-
2.2.6542
5.13×
4.184 =
5.7236
0.2888J - mol⏺H=I
9.8205 kJ moΓ(2) T = 1
3.br>7.78 + 273.15 = 41
0.93K 蒸发热可按WatSon经验式讣算:I-T °”P)士竺=
0.641342
5.13=出生="42
5.13∏ =1
9.
8.05X(^)^ =8.0415 kJ5「(3)由 ClaPCyrOn 方程知,d严 _ △/TF T∆vCiT由数据拟合In 〃哪〜丄得到In//^ =2
1.752-2
8.8.4/Γ
8.
3.4x313.15 3.2217~
1.6× 10^4 -
2.985 × 10^5 ^
3.3.1505 × 1.6× 10^4 × (1.6× 10^4 + 2.985 × 10^5) =1.401 × IO7 Pa = 133 atm > p ⅛ = IOO atm故储气罐会发生爆炸。
2.1
6.乙烷是重要的化工原料,也可以作为冷冻剂。现装满290 K、2.
4. MPa乙烷蒸气的钢瓶,不小 心接近火源被加热至478 K,而钢瓶的安全工作压力为4.5 MPa,问钢瓶是否会发生爆炸?(用RK 方程计算)解:由附录査得乙烷的临界参数。Tc=3
0.br>5.
4., Pc=4.884MPa, Vc=
1.48× 104m3∕mol; g>=0.091) T=290K, P=2∙48Mpa∙/ Tr=T∕Tt∙=
0.95 Pr=P∕Pv=0.51・•・使用普遍化第二维里系数法。Bo-
0.083-0.422/0.9516 --0.375B#-
0.139-0.172/0.9542 --0.074«/>—■ B° + flrfΓ ■ -
0.375 + 0.098x(-0.074) ■ -0.3823/. ½ = :R = 1
6..74 cm3 - moΓ1 Piz2=z? + ωi1 =
0.675 + 0.225 × 0.2 = 0.720.-.V2 = 14
9.9 cn√.mol-故 M∕ = V<z-%=149∙9-166∙74 = -16∙84 cm3 mol l过程①用普遍化压缩因子法查图得 也工.=_2 34 I = -O 89RTC ■ "^7?.仝血+ q<
1.2∙54RA R7; R7;・•. H, = -
2.5
4.
8.314 × 304.2 = -6424 kJ - kmoΓ查图得:IAL-L = -140R^(Sr):(Srl_R R RSf= -
1.5
8.χ 8.
3.4 = -13.14kJ kmoΓ, - K1过程②C: =
2.br>6.75 +
4..258 XIO-ST+ (-14.25×1 O^)T2::;Cw27
9.6 KJ kmol-ASJ = f∙w∙,sSLJT-RIn=
3.635 kJ kmoΓ, K'1 P JWiS T O53过程③用普压法:= + ω^- = -
1.72-
0.225 X 0.20 = -1.78R7; R7; R7;.= !⅛
1.+ O LlLL = -
0.95 - 0.225 × 0.20 = -1.02R R R/. λ∕f =-
1.78 × 31
4.X 304.2 = -4501.8 kJ kmolSf=-
1.02×
8.3I4 = -8.48kJ kmol l ∙K1ΔH = (-H:) + (∕∕f) + ∆Hld= 6424 + 327
9.6-450
1.8 = 5202 kJ kmol'ΔS = (-Sf) +∆Sp,d +(Sf) = I
3.4 + 3.635 -
8.48 = 83 kJ - kmoΓ1 ∙ KjΔt∕ = ΔH-Δ∕7V =
5.02-115.7
9.x149.9一
8.053×219.3 = 4601 kJ kmoll⏺∆∕y
2.17(ln P-I)X
4.184Tap (
0.930-7;,,)・..Tll =
2.
3.15 ÷ (-
0.5) = 272.65 K由表中数据知,正丁烷的临界压力,临界温度,如下:P=
3.786MPa = 38atmΓ =42
5.13K⏺
2.1
7.ln 37.86-1)
0.93-
2.2.6542
5.13×
4.184 =
5.7236
0.2888J - mol⏺H=I
9.8205 kJ moΓ(2) T = 1
3.br>7.78 + 273.15 = 41
0.93K 蒸发热可按WatSon经验式讣算:I-T °”P)士竺=
0.641342
5.13=出生="42
5.13∏ =1
9.
8.05X(^)^ =8.0415 kJ5「(3)由 ClaPCyrOn 方程知,d严 _ △/TF T∆vCiT由数据拟合In 〃哪〜丄得到In//^ =2
1.752-2
8.8.4/Γ
题目解答
答案
正确