any one know how i can solve this question?
What are the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the following constant pressure process for a system containing 0.956 moles of CH3OH ?
CH3OH(g, 122.0 ºC, 1.00 atm) ⟶ CH3OH(l, 34.0 ºC, 1.00 atm)
Assume that CH3OH(g) behaves as an ideal gas and the volume of CH3OH(l) is much less than that of CH3OH(g). Also, assume that the temperature of the surroundings is 34.0 ºC.
Data:
Molar heat capacity for CH3OH(g), Cp,m = 44.1 J K−1 mol−1
Molar heat capacity for CH3OH(l), Cp,m = 81.1 J K−1 mol−1
Enthalpy of vaporization, ΔvapH = 35.2 kJ mol−1 at 64.7 ºC and 1.00 atm
This may get you started. What is the boiling point for CH3OH? I found 64.7 C.
For q, you are moving gaseous CH3OH from 122 C to 64.7, then you condense it, then you move the liquid from 64.7 C to 34.0 C. Here are the three steps.
q1 = moles x Cp(g) x (T2-T1) = 0.956 x 44.1 x (122.0 - 64.7) = ?for gas
q2 = dHvap x mols = ? for condensation
q3 = moles x Cp(l) x (64.7 - 34.0) = ? for liquid
q total = q1 + q2 + q3
Work. That is -p(delta V) or = p(V2-V1)
Use PV = nRT to solve V of the gas. That is V1. The problem tells you to ignore V2(the liquid volume).
To solve this question, you need to calculate the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the given constant pressure process. Here are the steps to find each value:
1. Calculate q (heat transferred):
q = nCΔT
Where n is the number of moles of CH3OH, C is the molar heat capacity, and ΔT is the change in temperature. Since the process is at constant pressure, the heat transferred (q) is equal to the change in enthalpy (ΔH).
For the gas phase:
q(g) = n(g)Cp,m(g)(Tfinal(g) - Tinitial(g))
q(g) = 0.956 mol * 44.1 J K^(-1) mol^(-1) * (34.0 ºC - 122.0 ºC)
For the liquid phase:
q(l) = n(l)Cp,m(l)(Tfinal(l) - Tinitial(l))
q(l) = 0.956 mol * 81.1 J K^(-1) mol^(-1) * (34.0 ºC - 1.00 ºC)
2. Calculate w (work done):
The work done during this process can be considered negligible since the volume of the liquid phase is stated to be much less than that of the gas phase. Therefore, w ≈ 0.
3. Calculate ΔU (change in internal energy):
Since the process is at constant pressure, ΔU is equal to the heat transferred (q).
ΔU = q = q(g) + q(l)
4. Calculate ΔH (change in enthalpy):
ΔH = ΔU + PΔV
Since the process is at constant pressure, PΔV = w and w ≈ 0. Therefore, ΔH ≈ ΔU.
ΔH = ΔU = q(g) + q(l)
5. Calculate ΔS (change in entropy):
The change in entropy can be calculated using the formula:
ΔS = ΔH / T
Where ΔH is the change in enthalpy and T is the temperature.
For the gas phase:
ΔS(g) = ΔH(g) / T(g)
ΔS(g) = (q(g) + q(l)) / T(g)
For the liquid phase:
ΔS(l) = ΔH(l) / T(l)
ΔS(l) = (q(g) + q(l)) / (T(l) - T(g))
6. Calculate ΔSsurr (change in entropy of the surroundings):
Since the temperature of the surroundings is given as 34.0 ºC, ΔSsurr = (-ΔH) / T(surr)
7. Calculate ΔSuniv (change in total entropy):
ΔSuniv = ΔS(g) + ΔS(l) + ΔSsurr
Now, substitute the given values into the respective formulas and calculate each value step-by-step.