lecture 8

lecture 8 - More examples of the 2nd law: 2nd law...

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Unformatted text preview: More examples of the 2nd law: 2nd law and heat engines: It is impossible to take heat q from a reservoir, convert it all into work, and return into the original state (no cyclically operating heat engine with an efficiency equal to 1). It is however possible to design a cyclic heat engine that takes some heat Q1 from a warmer reservoir (T1), gives heat to a colder reservoir (T2), and produces mechanical work that is less than Q1. One such example is given in HW2. Irreversible vs reversible expansion of a gas. Consider a mole of an ideal gas occupying half a vessel whose total volume is V. Now we break the wall separating the gas from the empty part of the vessel so that the gas spreads over the entire vessel. Assume that there’s no heat exchange with the environment. Then the total energy is conserved and so the temperature of the gas is unchanged. The change in the entropy is then given by R lnV(2) – R ln V(1) = R ln (V(2)/V(1)) = R ln 2. The entropy has increased. Note that the entropy of the entire world has increased because we haven’t done anything to the environment and so its entropy did not change. Now consider a different experiment: perform reversible isothermal expansion of the gas from volume V/2 to V. The entropy change of the gas is precisely the same as before (R ln 2) because it is a function of state. However the entropy change of the environment is negative! To see this, note that the environment has done a negative work in the isothermal expansion, which is equal to (we have derived it in a lecture) W =  ­RT ln (V(2)/V(1)) =  ­RT ln 2. As the internal energy of the gas has not changed, the heat received by the gas is positive and equal to –W = RT 2. The heat received by the environment has the same absolute value but is negative and equal to q = –RT 2. And so the change of the environment’s entropy is q/T =  ­R ln 2. So we discover that the total change in the entropy of the world as a result of this process is zero, as expected for a reversible process. Mixing of two gases Let’s say we have n1 moles of one gas and n2 moles of another gas, both at the same pressure and temperature, P and T, and separated by a partition in the same insulated vessel. Now we break the partition and let them mix. Assume that both are ideal gases. Then P V1 = n1 RT, PV2 = n2 RT. After the gases mix we have P(V1 + V2) = (n1 + n2) RT so that V1/(V1 + V2) = n1/(n1 + n2) and so on. Using the expression for the entropy change of an ideal gas, we obtain ΔS = n1 R(ln[(V1 + V2)/V1 ] +n2 ln[(V1 + V2)/V2 ]) = R(n1 ln[(n1 + n2)/n1 ] + n2 ln[(n1 + n2)/n2 ]) > 0 Question: what if the two gases are identical? Then nothing happens when the partition is opened but our formula predicts a change in the entropy! Need quantum mechanics to resolve this paradox (the Gibbs paradox). ...
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