Question

Question 12: (1 point)

What are the values of  q, w, ΔU, ΔH, ΔS, ΔS_surr, and ΔS_univ for the following a constant pressure process for a system containing 0.572 moles of CH₃OH ?

CH₃OH(l, 26.0 ºC, 1.00 atm)  ⟶  CH₃OH(g, 118.0 ºC, 1.00 atm)

Assume that the volume of CH₃OH(l) is much less than that of CH₃OH(g) and that CH₃OH(g) behaves as an ideal gas. Also, assume that the temperature of the surroundings is 118.0 ºC.

Data:

Molar heat capacity for CH₃OH(l), C_p,m = 81.1 J K⁻¹ mol⁻¹

Molar heat capacity for CH₃OH(g), C_p,m = 44.1 J K⁻¹ mol⁻¹

Enthalpy of vaporization, Δ_vapH = 35.2 kJ mol⁻¹ at 64.7 ºC and 1.00 atm

Enter answers that are accurate to three (3) or more significant figures.

q = ____________ J

w = ____________ J

ΔU = ____________ J

ΔH = ____________ J

ΔS = ____________ J K⁻¹

ΔS_surr = ____________ J K⁻¹

ΔS_univ = ____________ J K⁻¹

Answer 1

CH3OH(g, 121.0 ºC, 1.00 atm) ⟶ CH3OH(l, 21.0 ºC, 1.00 atm)

The process is happening at a constant pressure (isobaric process) and CH3OH (g) is behaving as an ideal gas. The process is happening in three steps:

(i) cooling of the CH3OH (g) from 121.0 ºC to the 64.7 ºC,

(ii) phase transition of CH3OH(g) to CH3OH(l) at 64.7 ºC and

(iii) cooling of CH3OH(l) from 64.7 ºC to 21 ºC. We will see all the steps separately and combine them together to see the net change in the thermodynamic parameters. Please note that the boiling point of CH3OH is 64.7 ºC as it appears from the question.

(i) cooling of the CH3OH (g) from 121.0 ºC to the 64.7 ºC

The heat taken away from the ideal gas system in a isobaric process,

q=Delta H=nC_{p}Delta T .... (1.0)

where n = number of moles of the system, Cp is molar specific heat capacity at constant pressure and Delta T = difference in temperature.

Here, n = 0.195 mol, Cp = 44.1 JK-1mol-1, Delta T = (64.7-121.0) ºC = -56.3 ºC = -56.3 K

therefore q=Delta H = (0.195 mol) * (44.1 JK-1mol-1) (-56.3 K) = -484.15 J

Negative sign indicates that the heat is taken away from the system. That is, the process is exothermic.

The change in internal energy is independent of pressure and it can be calculated as below for an ideal gas,

NJ = nC,AT. . . . (1.1)

where Cv is molar specific heat capacity

Now, we know that, for an ideal gas,

Cp - Cv = R ...... (1.2)

Where R is universal gas constant = 8.314 JK-1mol-1

Cv = Cp - R = 44.1 JK-1mol-1 - 8.314 JK-1mol-1 = 35.786 JK-1mol-1

therefore Delta U= (0.195 mol) * (35.786 JK-1mol-1) (-56.3 K) = -392.877 J

Negative sign indicates that the internal energy of the system is decreasing.

For a reversible heating/cooling process at constant P (reversible isobaric), the entropy change of the system can be calculated as follow,

deltas5.gif..... (1.3)

Delta S= (0.195 mol) (44.1 JK-1mol-1) ln ((64.7 K)/(121 K)) = (0.195 mol) (44.1 JK-1mol-1) (-0.626) = -5.384 JK-1

Negative sign indicates that the entropy of the system is decreasing.

For a reversible process Delta S=-Delta S_{surr} = 5.384 JK-1

and

Delta S_{uni}=Delta S+Delta S_{surr}=0

ii) phase transition of CH3OH(g) to CH3OH(l) at 64.7 ºC

For phase transition,

q=Delta H=n Delta condH .... (1.4)

where Delta condH = molar latent heat of condensation = -Delta vap H = -35.2 kJ mol-1 therefore q=Delta H = (0.195 mol) * (-35.2 kJ mol-1) = - 6.864 kJ

Delta S =Delta condH/T ... (1.5)

therefore Delta S = (-35.2 kJ mol-1)/((64.7+273) K) = (-35.2 kJ mol-1)/(337.7 K) = -104.235 JK-1

therefore Delta S_{surr} = 104.235 JK-1

Delta S_{uni}=0

(iii) cooling of CH3OH(l) from 64.7 ºC to 21 ºC

q=Delta H=n C Delta T .... (1.6)

C = molar specific heat capacity of liquid CH3OH = 81.1 J K−1 mol−1

delta T = (21-64.7) ºC = -43.7 ºC = -43.7 K

therefore q=Delta H = (0.195 mol) * (81.1 J K−1 mol−1) (-43.7 K) = -681.094 J

For a liquid change in volume on cooling is almost 0. So, the change in internal energy will be equal to the q.Delta S= n Clnrac {T_{2}}{T_{1}} = (0.195 mol) (81.1 JK-1mol-1) ln ((21 K)/(64.7 K)) = (0.195 mol) (81.1 JK-1mol-1) (-1.125) = -17.795 JK-1

therefore Delta S_{surr} = 17.795 JK-1

Delta S_{uni}=0

Now one can sum over all the values from three steps to get the net value of q, delta H, delta U, deltaS and other thermodynamic parameter. Note that changes happened during the cooling have been considered here as the reversible process to calculate the thermodynamic parameters. For non reversible process thermodynamic parameter for ideal gases will differ. However, getting the value of entropy is a tedious task for a nonreversible process.

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