A 2.00-L container has a center partition that divides it into two equal parts a
ID: 2130080 • Letter: A
Question
A 2.00-L container has a center partition that divides it into two equal parts as shown in Figure P22.41. The left side contains 0.044 0 mol of H2 gas, and the right side contains 0.044 0 mol of O2 gas. Both gases are at room temperature and at atmospheric pressure. The partition is removed, and the gases are allowed to mix. What is the entropy increase of the system?
A 2.00-L container has a center partition that divides it into two equal parts as shown in Figure P22.41. The left side contains 0.044 0 mol of H2 gas, and the right side contains 0.044 0 mol of O2 gas. Both gases are at room temperature and at atmospheric pressure. The partition is removed, and the gases are allowed to mix. What is the entropy increase of the system?Explanation / Answer
The ideal entropy of mixing (and we will assume these gases behave ideally) of two distinguishable chemical components that are initially separated and at the same temperature and pressure is given by:
?Smixing = -n*R*[x1*ln(x1) + x2*ln(x2)]
where n is the total number of moles in the mixture, x1 is the mole fraction of component 1 and x2 is the mole fraction of component 2. R, of course, is the universal gas constant.
Here, n = (0.064+0.056)mol = 0.12mol, x1 = 0.064/0.12 =0.533 and x2 = 0.056/0.12 = 0.467, so:
?Smixing = -(0.12mol)*R*[0.533*ln(0.533) + 0.467*ln(0.467)]
?Smixing = 0.083*R*mol = 0.6507 J/K
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