PHYS6562 W5 Daily Question

Entropy of socks

For a child’s room containing 100 things in the room, we only need to consider the configuration space since those things are stationary in the room. The room before tightening up, the number of configurations is

\[\Omega = V^N = 5^{100}\]

Let’s consider the center-of-mass location accuracy of 1 cm. First of all, the center-of-mass can be located itself at any point of the room. Therefore,

\[\Omega_{CM} = 5\]

Then we can assume that the center of mass in \(x\) direction is at \(0\). For each randomly located thing \(A1\) at \(x\), there would be the another thing \(A1'\) located at the \(-x\). We now have only 50 things we can throw at will. For the \(A50'\), since the deviation of 1cm is acceptable, \(A50'\) can have roughly 1m of freedom in the \(x\) direction. Therefore,

\[\Omega_{item} = (L^{50}*1)^3 = 5^{50}\]

where \(L = 5^{1/3}\).

\[\Omega' = \Omega_{CM}\times\Omega_{item} = 5^{51}\]

Entropy decrease can be found to be

\[\Delta S = k_B\log{\Omega'} - k_B\log{\Omega} = -49k_B\log{5}\]

\[\Delta S_{total} = 10^9\times\Delta S = 1.09\times10^{-12}\]

For 1 liter baloon contracting by 1% in volume, \[\Delta S = Nk_B\log{0.99V} - Nk_B\log{V} = \frac{PV}{T}\log(0.99)\] where \(P = 1atm\), \(V = 1L\) and \(T = 300K\). \[\Delta S = -0.0034 (J/K)\]