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 $27^{°} C$ (if all gases behave ideally) is $(R = 2 cal {mol}^{- 1} K^{- 1})$ .

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Solution

The heat of combustion of a substance, also known as the calorific value or the energy value, can be defined as the amount of heat liberated when a given amount of the substance undergoes combustion.
Usually, the heat of combustion is considered to be a synonym of calorific value, which can be defined as the total amount of energy liberated when a given mass of a substance undergoes complete combustion in the presence of (an adequate quantity of) oxygen under standard conditions for temperature and pressure.
The heat of combustion of Ethanol :
$C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2 {CO}_{2} (g) + 3 H_{2} O (l)$
Solution: Since all gases behave ideally, they must follow the ideal gas equation i.e., $PV = nRT$ .
And we have to find out the value of $∆_{c} U$ , so we can use the equation,
$∆_{c} H = ∆_{c} U + PV$ , but we are not given with pressure and volume of the gas, so by using the ideal gas equation we can replace $PV$ with $nRT$ .
$∆ n_{g} = Number of moles of gases on (product side - reactant side) = 2 - 3 = - 1 ∆_{c} H = ∆_{c} U + (∆ n_{g}) RT ∆_{c} H = ∆_{c} U - RT ∆_{c} U = ∆_{c} H + RT = - 327 \times 10^{3} + 2 \times 300 = - 326400 cal .$
Therefore, the heat evolved at constant volume and $27^{°} C$ is $- 326400 cal$ .

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

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 $27^{°} C$ (if all gases behave ideally) is $(R = 2 cal {mol}^{- 1} K^{- 1})$ .