Physical Chemistry: Thermodynamics, Structure, and Change

Published by W. H. Freeman
ISBN 10: 1429290196
ISBN 13: 978-1-42929-019-7

Chapter 2 - Topic 2B - Enthalpy - Exercises - Page 104: 2B.1(b)

Answer

$$C_{p,m} = 52.6 \space J \space K^{-1} \space mol^{-1}$$ $$C_{v,m} = 44.3 \space J \space K^{-1} \space mol^{-1}$$

Work Step by Step

$$\Delta H = C_{p} \Delta T$$ $$\frac{\Delta H}{\Delta T} = C_p$$ $$C_p = \frac{178 \space J}{1.78 \space K} =100 \space J \space K^{-1}$$ $$C_{p,m} = \frac{C_p}{n } = \frac{100 \space J \space K^{-1}}{1.9 \space mol} = 52.6 \space J \space K^{-1} \space mol^{-1}$$ $$C_{p,m} - C_{v,m} = R$$ $$C_{v,m} = C_{p,m} - R$$ $$C_{v,m} = 52.6 \space J \space K^{-1} \space mol^{-1} - 8.314 \space J \space K^{-1} \space mol^{-1}$$ $$C_{v,m} = 44.3 \space J \space K^{-1} \space mol^{-1}$$
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