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# According to the Sackur-Tetrode equation, the entropy of a ISBN: 9780201380279 40

## Solution for problem 35P Chapter 2

An Introduction to Thermal Physics | 1st Edition

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Problem 35P

According to the Sackur-Tetrode equation, the entropy of a monatomic ideal gas can become negative when its temperature (and hence its energy) is sufficiently low. Of course this is absurd, so the Sackur-Tetrode equation must be invalid at very low temperatures. Suppose you start with a sample of helium at room temperature and atmospheric pressure, then lower the temperature holding the density fixed. Pretend that the helium remains a gas and does not liquefy. Below what temperature would the Sackur-Tetrode equation predict that S is negative? (The behavior of gases at very low temperatures is the main subject of Chapter 7.)

Step-by-Step Solution:

Step 1 of 5</p>

The Sackur Tetrode equation gives the entropy of a monatomic gas molecule as, ……….1

Where S is entory, N is avogadro number, k and h are constants, m is mass of argon atom, V is volume and U is internal energy.

Step 2 of 5</p>

In the above equation, if the internal energy U drops low enough such that the log term can go below making the entropy S negative. So the Sackur-Tetrode equation breaks down at lower energies. This implies that, at lower temperatures things goes different.

From equation 1, Using    ………….2

Step 3 of 5</p>

For example, suppose we have a mole of helium and we cool it (assuming it remains as gas).

Then the critical temperature is given by,

Rearranging equation 2 for temperature,    ……...3

Therefore, the critical temperature is the temperature below which the entropy will be negative.

Step 4 of 5

Step 5 of 5

##### ISBN: 9780201380279

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