The molar heat capacity (C$$_p$$) of CD$$_2$$O is 10 cals at 1000 K. The change of entropy associated with cooling of 32 g of CD$$_2$$O vapour from 1000 K to 100 K at constant pressure will be: (D = deuterium, atomic mass = 2u)

The molar heat capacity at constant pressure, $$ C_p $$, for CD$$_2$$O is given as 10 cal/(mol·K) at 1000 K. We need to find the change in entropy ($$\Delta S$$) when cooling 32 g of CD$$_2$$O vapor from 1000 K to 100 K at constant pressure.

The entropy change at constant pressure is given by the integral $$\Delta S = \int \frac{dQ_{\text{rev}}}{T}$$. For a reversible process at constant pressure, $$dQ_{\text{rev}} = n C_p dT$$, where $$n$$ is the number of moles and $$C_p$$ is the molar heat capacity. Thus, $$\Delta S = \int_{T_1}^{T_2} \frac{n C_p}{T} dT$$.

Since $$C_p$$ is provided as 10 cal/(mol·K) at 1000 K and no temperature dependence is specified, we assume $$C_p$$ is constant over the temperature range from 100 K to 1000 K. Therefore, the equation simplifies to $$\Delta S = n C_p \int_{T_1}^{T_2} \frac{dT}{T} = n C_p \ln\left(\frac{T_2}{T_1}\right)$$.

Here, the initial temperature $$T_1 = 1000$$ K and the final temperature $$T_2 = 100$$ K, so $$\frac{T_2}{T_1} = \frac{100}{1000} = 0.1$$. The natural logarithm of 0.1 is $$\ln(0.1) = \ln\left(\frac{1}{10}\right) = -\ln(10)$$. Using $$\ln(10) \approx 2.302585$$, we get $$\ln(0.1) \approx -2.302585$$.

Now, we need the number of moles $$n$$. The molar mass of CD$$_2$$O is calculated as follows: carbon (C) has atomic mass 12 u, deuterium (D) has atomic mass 2 u (given), and oxygen (O) has atomic mass 16 u. So, molar mass = 12 + 2 \times 2 + 16 = 12 + 4 + 16 = 32 g/mol.

The mass given is 32 g, so $$n = \frac{\text{mass}}{\text{molar mass}} = \frac{32}{32} = 1$$ mole.

Substituting the values: $$n = 1$$ mol, $$C_p = 10$$ cal/(mol·K), and $$\ln\left(\frac{T_2}{T_1}\right) = -2.302585$$, we get:

$$\Delta S = 1 \times 10 \times (-2.302585) = 10 \times (-2.302585) = -23.02585$$ cal/K.

Rounding to two decimal places, $$\Delta S \approx -23.03$$ cal K$$^{-1}$$ (or cal deg$$^{-1}$$).

The negative sign indicates a decrease in entropy due to cooling. Comparing with the options:

• A. 23.03 cal deg$$^{-1}$$
• B. $$-23.03$$ cal deg$$^{-1}$$
• C. 2.303 cal deg$$^{-1}$$
• D. $$-2.303$$ cal deg$$^{-1}$$

Option B matches the calculated $$\Delta S$$. Hence, the correct answer is Option B.