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
q = -35.2 kJ
w = 0 kJ
ΔU = -35.2 kJ
ΔH = -35.2 kJ
ΔS = -44.1 J K−1 mol−1
ΔSsurr = 0 J K−1 mol−1
ΔSuniv = -44.1 J K−1 mol−1
To determine the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the given constant pressure process, we can use various thermodynamic equations and principles. Let's break down each value.
1. q (heat):
Heat transfer (q) is given by the equation:
q = n * Cp * ΔT
where n is the number of moles, Cp is the molar heat capacity, and ΔT is the change in temperature. In this process, there is a temperature change from 122.0 ºC to 34.0 ºC. Since CH3OH is in the gas phase initially and then changes to the liquid phase, there are multiple steps involved.
a) First, calculate the heat of vaporization (ΔHvap):
ΔHvap = ΔvapH * n
where ΔvapH is the enthalpy of vaporization. In this case, ΔvapH = 35.2 kJ mol−1.
b) Next, calculate the heat during the vaporization process:
q_vap = ΔHvap
This step only involves the enthalpy change during vaporization.
c) Finally, calculate the heat during the cooling process:
q_cool = n * Cp,l * ΔT
where Cp,l is the molar heat capacity of the liquid phase. In this case, Cp,l = 81.1 J K−1 mol−1.
The total heat (q) can be computed by summing the values from steps b and c.
2. w (work):
For a constant-pressure process, the work done (w) is given by:
w = −PΔV
Since ΔV is not given explicitly, we need to use the ideal gas law to calculate the volume change.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J K−1 mol−1), and T is the temperature. From this equation, we can rearrange it to solve for V:
V = nRT / P
We can then calculate the volume change (ΔV) from the initial and final conditions.
3. ΔU (change in internal energy):
The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = q + w
Substitute the values of q and w calculated earlier to determine ΔU.
4. ΔH (change in enthalpy):
The change in enthalpy (ΔH) is related to the change in internal energy and work done:
ΔH = ΔU + PΔV
The values for ΔU and ΔV can be determined from the previous calculations.
5. ΔS (change in entropy):
To calculate the change in entropy (ΔS), we need the molar entropies of the initial and final states:
ΔS = ∫(Cp/T) dT
Integrate the molar heat capacity (Cp) with respect to temperature (T) from the initial to final temperature conditions.
6. ΔSsurr (change in entropy of the surroundings):
For a constant-temperature process, the change in entropy of the surroundings (ΔSsurr) can be defined as:
ΔSsurr = -q / T_surr
where T_surr is the temperature of the surroundings. Substituting the value of q calculated earlier will give you ΔSsurr.
7. ΔSuniv (change in total entropy):
The change in total entropy (ΔSuniv) can be evaluated using the equation:
ΔSuniv = ΔS + ΔSsurr
Add the values of ΔS and ΔSsurr calculated earlier to determine ΔSuniv.
By following these steps, you should be able to obtain the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the given process involving CH3OH.
To find the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the given process, we'll go step-by-step:
Step 1: Calculate the heat transferred (q) for the process.
q = nCpΔT
Given:
n = 0.956 moles (number of moles of CH3OH)
Cp = 44.1 J K^(-1) mol^(-1) (molar heat capacity for CH3OH(g))
ΔT = (34.0 °C - 122.0 °C) = -88 °C = -88 K (temperature change)
Substituting the values:
q = 0.956 mol × 44.1 J K^(-1) mol^(-1) × (-88 K)
q = -3675.81 J
Therefore, the value of q is -3675.81 J.
Step 2: Calculate the work done (w) for the process.
w = -PΔV
Given:
P = 1.00 atm (pressure)
ΔV = Since the volume of CH3OH(l) is much less than that of CH3OH(g), we can assume that the volume change is negligible.
Since ΔV is negligible, the work done (w) is also negligible.
Therefore, the value of w is approximately 0 J.
Step 3: Calculate the change in internal energy (ΔU).
ΔU = q + w
Substituting the values:
ΔU = -3675.81 J + 0 J
ΔU = -3675.81 J
Therefore, the value of ΔU is -3675.81 J.
Step 4: Calculate the change in enthalpy (ΔH).
ΔH = ΔU + PΔV
Given:
P = 1.00 atm (pressure)
ΔV = Since the volume of CH3OH(l) is much less than that of CH3OH(g), we can assume that the volume change is negligible.
Since ΔV is negligible, ΔH = ΔU.
Therefore, the value of ΔH is -3675.81 J.
Step 5: Calculate the change in entropy (ΔS).
ΔS = nCp,m(ln(T₂/T₁))
Given:
n = 0.956 moles (number of moles of CH3OH)
Cp,m = 44.1 J K^(-1) mol^(-1) (molar heat capacity for CH3OH(g))
T₁ = 122.0 °C = 395.15 K (initial temperature)
T₂ = 34.0 °C = 307.15 K (final temperature)
Substituting the values:
ΔS = 0.956 mol × 44.1 J K^(-1) mol^(-1) × ln(307.15 K / 395.15 K)
ΔS ≈ -3.12 J K^(-1)
Therefore, the value of ΔS is approximately -3.12 J K^(-1).
Step 6: Calculate the change in entropy of the surroundings (ΔSsurr).
ΔSsurr = -ΔH/T
ΔSsurr = (-1) × (-3675.81 J) / 307.15 K
Substituting the values:
ΔSsurr ≈ 11.97 J K^(-1)
Therefore, the value of ΔSsurr is approximately 11.97 J K^(-1).
Step 7: Calculate the total change in entropy (ΔSuniv).
ΔSuniv = ΔS + ΔSsurr
Substituting the values:
ΔSuniv ≈ -3.12 J K^(-1) + 11.97 J K^(-1)
ΔSuniv ≈ 8.85 J K^(-1)
Therefore, the value of ΔSuniv is approximately 8.85 J K^(-1).
To summarize:
q = -3675.81 J
w ≈ 0 J
ΔU = -3675.81 J
ΔH = -3675.81 J
ΔS ≈ -3.12 J K^(-1)
ΔSsurr ≈ 11.97 J K^(-1)
ΔSuniv ≈ 8.85 J K^(-1)