free energy ideal gas
Physics..quick problem. An ideal gas expands into a vacuum in a rigid container (free expansion).?
A) An increase of entropy of the gas
B) Increase in pressure of the gas
C) Increase of Temperature of the gas
D)decrease of temperature of the gas
E) A decrease of the internal energy of the gas
F) Both D and E
A.
Since this is an ideal gas, the temperature does not change at all during a free expansion. Because it’s expanding in a vacuum, it’s not doing any work to push other gases out of the way. Temperature/Internal energy is dependent on the heat added and work done by the gas. There is no heat added and no work done by the gas; thus, temperature/internal energy is constant. An increase in pressure makes no sense either, since volume is increased and temperature stays the same.
When the gas is expanding freely, it’s actually increasing its entropy. Entropy is a measure of the disorder of a system. You would expect the gas to spread out and become more disorderly instead of staying in one corner, clumped neatly and not spreading out.
From a mathematical perspective, we know that change in entropy is equal to heat added over the temperature, dS = dQ/T. Since change in internal energy is 0, dQ = dW (work) = P*dV. To find the change in entropy, we integrate dQ/T. We can factor out T because it is a constant. By subbing in P*dV in place of dQ, we can integrate with respect to volume. P = nRT/V so we end up with:
nR* integral of dV/V, which is n*R*ln(Vfinal/Vinitial).
Thus, change in entropy = n*R*ln(Vf/Vi)
Since Vfinal is greater than Vinitial, the change in entropy is positive, which signifies an increase of entropy in the gas.
Lecture 4 | Modern Physics: Statistical Mechanics
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