Question

The molar heat capacity (${{C}_{p}}$) of $C{{D}_{2}}O$ is 10 cals at 1000K. The change in entropy associated with cooling of 32g of $C{{D}_{2}}O$ vapour from 1000K to 100K at constant pressure will be: (D= deuterium, at mass=2u)
A. $-2.303\text{ }cal\text{ }de{{g}^{-1}}$
B. $-23.03\text{ }cal\text{ }de{{g}^{-1}}$
C. $23.03\text{ }cal\text{ }de{{g}^{-1}}$
D. $2.303\text{ }cal\text{ }de{{g}^{-1}}$

Answer 1

Hint: Note the values which are given in the question, then use the formula of the change of entropy when temperature changes, given the mass and molar heat capacity at constant pressure.

Complete step by step solution:
Let us first know what entropy is. Entropy, in thermodynamics, refers to the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. As the work is obtained from ordered molecular motion, we can also relate entropy as randomness. Hence, entropy is the degree of uncertainty and randomness in a system. As the temperature changes, the kinetic energy of the system decreases and hence, the randomness or entropy also decreases. We can write the equation for change in entropy as:
\[\mathbf{\Delta }S\text{ }=\text{ }2.303\times n\times {{C}_{p}}\times log(\dfrac{{{T}_{2}}}{{{T}_{1}}})\]
Here, ΔS is the change in entropy, ${{C}_{p}}$ is the molar heat capacity at constant pressure, T2 is the final temperature and T1 is the initial temperature.
Substituting values from the question, we get
\[\mathbf{\Delta }S=2.303\times (\dfrac{32}{32})\times 10\text{ }log(\dfrac{100}{1000})\], mass is 32g
Which comes out to be $-23.03\,cal\,{{\deg }^{-1}}$.

So, the change in entropy by changing temperature from 1000K to 100K is$-23.03\,cal\,{{\deg }^{-1}}$.

Note: Entropy is directly proportional to the temperature. It is so because as the temperature increases, the energy associated with the molecules increases and they collide more and overall, their randomness or uncertainty increases too, and vice- versa. That is why in the question, we have the change in entropy as negative. Temperature is decreasing from 1000K to 100K and so is the entropy. Also, check for calculation errors involving logarithm.