Consider a process in which an ideal gas is compressed to one-sixth of its original volume at constant temperature.
Calculate the entropy change per mole of gas.
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To calculate the entropy change per mole of gas during the process, we need to use the equation:
ΔS = nR ln(V2/V1)
where ΔS is the entropy change, n is the number of moles of gas, R is the ideal gas constant, V1 is the initial volume, and V2 is the final volume.
In this case, since we are given that the gas is compressed to one-sixth of its original volume, we can say that V2 = V1/6.
Substituting this into the equation, we have:
ΔS = nR ln((V1/6)/V1)
Simplifying further:
ΔS = nR ln(1/6)
To evaluate ln(1/6), we can use the property of natural logarithms that ln(a/b) = ln(a) - ln(b). In this case, we have:
ln(1/6) = ln(1) - ln(6)
Now, ln(1) = 0, so we can simplify further:
ln(1/6) = -ln(6)
Therefore, the entropy change per mole of gas is:
ΔS = - nR ln(6)