Suppose that 120 g of ethanol at 24◦C is mixed with 275 g of ethanol at 81◦C at constant atmospheric pressure in a thermally insulated vessel. What is the ∆Ssys for the process? The specific heat capacity for ethanol is 2.42J/g K. Answer in units of J/K.
Can you please explain to me how to solve this problem?
To solve this problem, you need to calculate the change in entropy (∆Ssys) for the process.
The change in entropy (∆Ssys) can be calculated using the equation:
∆Ssys = ∆Ssur + ∆Ssys
where ∆Ssur is the change in entropy of the surrounding, and ∆Ssys is the change in entropy of the system.
In this case, the process is happening at constant atmospheric pressure, which means there is no pressure-volume work involved. Therefore, the ∆Ssur can be assumed to be zero (∆Ssur = 0).
Now, the key is to calculate the change in entropy of the system (∆Ssys). The change in entropy can be calculated using the equation:
∆Ssys = nC∆T
where n is the number of moles, C is the molar heat capacity, and ∆T is the change in temperature.
To calculate n, you need to convert the masses of ethanol to moles using their molar mass. The molar mass of ethanol (C2H5OH) is approximately 46 g/mol.
n1 = Mass1 / Molar Mass
n2 = Mass2 / Molar Mass
n1 = 120 g / 46 g/mol ≈ 2.61 mol
n2 = 275 g / 46 g/mol ≈ 5.98 mol
Next, calculate the change in temperature (∆T) using the initial and final temperatures:
∆T = Final Temperature - Initial Temperature
∆T = 81◦C - 24◦C = 57◦C
Now, substitute the values into the equation:
∆Ssys = (n1 + n2) × C × ∆T
∆Ssys = (2.61 mol + 5.98 mol) × 2.42 J/g K × 57◦C
Finally, calculate the ∆Ssys:
∆Ssys = 8.59 mol × 2.42 J/g K × 57 K
∆Ssys ≈ 1120 J/K
Therefore, the change in entropy (∆Ssys) for the process is approximately 1120 J/K.