温度为300K,压力为101325pa的理想气体0.5mol,等温可逆膨胀到30L,再恒容升温至600K,求该过程中的W、Q、ΔH、ΔH、ΔS(已知Cv=20.79J/k.mol)需要详细分析过程 谢谢!
T0=300K,p0=101325Pa,算出V0=12.3L
T1=300K,V1=30L
T2=600K,V2=30L
整个过程分成了两步,分别计算,然后叠加
等温可逆膨胀过程:
dW = -pdV ==> W = -nRTln(V1/V0) = -1111J
温度不变 ==> ΔU = ΔH = 0
Q = ΔU - W = 1111J
ΔS = Q/T = 3.70J/K
恒容过程
W = 0
ΔU = Q = nCvΔT = 3119J
ΔH = ΔU + Δ(pV) = ΔU + nRΔT = 4366J
dS = dQ/T = nCvdT/T ==> ΔS = nCvln(T3/T2) = 7.21J/K
合计
W = -1111J
Q = 4230J
ΔU = 3119J
ΔH = 4366J
ΔS = 10.91J/K
这不是大学普通化学题,算是中级无机化学或物理化学了
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