Test03S-docA

# 4 4 3 4 4 3 a 134x10 pa b 610x10 pa c 820x10 pa

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Unformatted text preview: ou derive the equation used to calculate the change in chemical potential! The change in chemical potential with pressure for an ideal gas is given by: Δµ = RT ln (pF / pI) = 8.3145 J/mol.K x 360 K x ln (VI / VF) = 750 J/mol.K Question 12: (5 points) o Exactly two moles of substance Y in the liquid state are placed in an empty 0.5 L- container at - 20.5 C and the container is sealed and equilibrated. Given the thermodynamic data provided in the data section for substance Y, determine the equilibrium pressure of the vapor phase at this temperature. 4 4 3 4 4 3 a) 1.34x10 Pa b) 6.10x10 Pa c) 8.20x10 Pa d) 9.41x10 Pa e) 2.56x10 Pa f) 1.10x10 Pa Show your calculations and state your assumptions! Here we use the result of question 18. ln (P2 / P1) = - (ΔvapH/R) (1/T2 – 1/T1) o Where P1 = 101325 Pa and T1 = - 9.2 C = 273.15 – 9.2 K = 263.95 K and T2 = - 20.5oC = 273.15 – 20.5 K = 252.65 K, ΔvapH = 24943 J/mol, R = 8.3145 J/mol.K 4 So, P2 = 6.10x10 Pa Question 13: (5 points) The state function whose change at constant temperature and pressure is defined as the maximum non- expansion work is equal to: a) U b) U+pV c) S d) PV e) U+pV- TS f) U- TS Justify your reasoning. The maximum non- expansion work at constant pressure and temperature is the change in Gibbs free energy for the process (ΔG). We know that G = H – TS = U + pV – TS. Question 14: (5 points) The temperature dependence of the vapor pressure of the liquid...
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