JEE Main 2014 · 11 Apr · Shift 1 · Chemistry Q39

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ChemistrySingle correct+4 / -1Question 39 of 90

The molar heat capacity ( $C_p$ ) of $\ce{CD2O}$ is 10 cals at 1000 K. The change in entropy associated with cooling of 32 g of $\ce{CD2O}$ vapour from 1000 K to 100 K at constant pressure will be :

(D = deuterium, at. mass = 2 u)

23.03 cal deg

^{-1}

-23.03

cal deg

^{-1}

2.303 cal deg

^{-1}

-2.303

cal deg

^{-1}

Answer and worked solution

Official answer: B

$M(\ce{CD2O})=12+4+16=32$ , so 32 g is 1 mol.
$\Delta S=nC_p\ln\frac{T_2}{T_1}=10\ln\frac{100}{1000}=-23.03\text{ cal deg}^{-1}.$

Topics:Entropy

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JEE Main 2014 · 11 Apr · Shift 1, Chemistry question 39