calculate the q, w, delta U, delta H and delta S.?
a sample consisting 3.0mol of diatomic perfect gas molecule at 200K is compressed reversibly and adiabatically until its temperature reaches 250K. Given that C v,m =275 J/K*mol, calculate q, w, delta U, delta H and delta S.
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Verified answer
n = 3 moles
T1 = 200 K
T2 = 250 K
Cv = 275 J/mol*K
Adiabatic reversible means Q = 0 and pv^k= C, k = Cp/Cv
Calculate Q
Q = 0
Calculate Work, W. Use energy balance,
∆U = Q - W, Q = 0
n*Cv*∆T = -W
W = -(3*275*[250-200])
W = - 41250 J = - 41.25 kJ (negative indicates process of compression)
Calculate ∆U,
∆U = -W = 41.25 kJ
Calculate ∆H
∆H = Cp∆T
For diatomic gas Cv = 5/2 R so Cp = 7/2 R
So, R = 2/5*Cv = 0.4*275 J/mol.K = 110 J/mol.K
Cp = 7/2R = 385 J/mol.K
Hence, ∆H = nCp∆T = 3*385*50 = 57750 J = 57.75 kJ
Calculate ∆S
∆S = n*[Cv ln (T2/T1) - R ln(V2/V1)]
Calculate V2/V1 first.
V2/V1 = (T2/T1)^1/(k-1), k=cp/cv = 1.4
V2/V1 = (250/200)^(1/0.4)
V2/V1 = 1.747
So,
∆S = 3*[275 ln (250/200) - 110*ln(1.747)]
∆S = 760.602 J/K = 0.761 kJ/K
Delta U Q-w
a million.) 681.2 kj/mol 2.) eighty.3 J/mol 3.) 665.14 kj/molK all 3 are the sum of goods minus the sum of reactants. yet because you're actually not given values for gibbs loose power you will discover it with the equation deltaG= deltaH- temp x deltaS. As for the sencond 0.5 of each and every question I ignore and that i dont opt to respond to incorrectly