Medium INTEGER +4 / -1 PYQ · JEE Mains 2020

The heat of combustion of ethanol into carbon dioxide and water is – 327 kcal at constant pressure. The heat evolved (in cal) at constant volume and 27oC (if all gases behave ideally) is (R = 2 cal mol–1 K–1) ________.

Answer (integer) 326400

Solution

C<sub>2</sub>H<sub>5</sub>OH (l) + 3O<sub>2</sub>(g) $\to$ 2CO<sub>2</sub>(g) + 3H<sub>2</sub>O(l) <br><br>$\Delta$H = –327 kcal; $\Delta$n<sub>g</sub> = – 1 <br><br>$\Delta$H = $\Delta$U + $\Delta$n<sub>g</sub>RT <br><br>$\Rightarrow$ $\Delta$U = – 327 + 2 × 10<sup>–3</sup> × 300 <br><br>= – 326.4 kcal <br><br>= – 326400 cal <br><br>$\therefore$ Heat evolved = 326400 cal.

About this question

Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law

This question is part of PrepWiser's free JEE Main question bank. 110 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →