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Chapter 20: Problem 59

If liquid nitrogen is boiled slowly-that is, reversiblyto transform it into nitrogen gas at a pressure \(P=100.0 \mathrm{kPa}\), its entropy increases by \(\Delta S=72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K}) .\) The latent heat of vaporization of nitrogen at its boiling temperature at this pressure is \(L_{\text {vap }}=5.568 \mathrm{~kJ} / \mathrm{mol}\). Using these data, calculate the boiling temperature of nitrogen at this pressure.

Short Answer

Expert verified
Answer: The boiling temperature of nitrogen at the given pressure is approximately 77.2 K.

Step by step solution

01

Rewrite the formula for Temperature

Given the formula \(\Delta S = \frac{L_{\text {vap}}}{T}\), we can rewrite the formula to find \(T\) by multiplying both sides with \(T\) and dividing by \(\Delta S\): \(T = \frac{L_{\text {vap}}}{\Delta S}\)
02

Substitute the values

Now, we can substitute the given values into the formula: \(T = \frac{5.568 \mathrm{~kJ} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})}\) First, we should convert the latent heat of vaporization from kJ to J. Since 1 kJ = 1000 J: \(L_{\text {vap}} = 5.568 \mathrm{~kJ} \times \frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}} = 5568 \mathrm{~J} / \mathrm{mol}\) Now, substitute the converted value: \(T = \frac{5568 \mathrm{~J} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})}\)
03

Calculate the boiling temperature

We can calculate the boiling temperature by dividing \(L_{\text {vap}}\) by \(\Delta S\): \(T = \frac{5568 \mathrm{~J} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})} = \frac{5568}{72.1} \mathrm{K} \approx 77.2 \mathrm{K}\) The boiling temperature of nitrogen at the given pressure is approximately 77.2 K.

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