The heat of combustion of ethanol into carbon dioxides and water is $$-327$$ Kcal at constant pressure. The heat evolved (in cal) at constant volume at 27$$°$$C (if all gases behave ideally) is $$(R = 2$$ cal mol$$^{-1}$$ K$$^{-1})$$ ___________.

We are told that the heat of combustion of ethanol measured at constant pressure equals the enthalpy change of reaction, so we write

$$\Delta H = -327\ \text{Kcal}\;=\;-327 \times 1000\ \text{cal} = -327000\ \text{cal}$$

To obtain the heat evolved at constant volume we need the internal-energy change $$\Delta U.$$

For any reaction involving ideal gases the relation between enthalpy change and internal-energy change is first stated:

$$\boxed{\;\Delta H = \Delta U + \Delta n_g R T\;}$$

Here $$\Delta n_g$$ is the difference (moles of gaseous products) $$-$$ (moles of gaseous reactants), $$R$$ is the gas constant and $$T$$ is the absolute temperature.

We now write the balanced combustion of ethanol with liquid water as product:

$$\mathrm{C_2H_5OH(l) + 3\,O_2(g) \;\longrightarrow\; 2\,CO_2(g) + 3\,H_2O(l)}$$

Only the gaseous species are counted for $$\Delta n_g$$. On the right we have $$2$$ mol $$CO_2(g)$$ and on the left $$3$$ mol $$O_2(g)$$. Therefore

$$\Delta n_g = 2 - 3 = -1$$

The temperature is $$27^{\circ}\mathrm{C} = 27 + 273 = 300\ \text{K}$$ and we are given $$R = 2\ \text{cal mol}^{-1}\text{K}^{-1}.$$

Substituting in the formula for $$\Delta U$$:

$$\Delta U = \Delta H - \Delta n_g R T$$

Because $$\Delta n_g = -1$$, the term $$-\Delta n_g$$ becomes $$-(-1) = +1,$$ so

$$\Delta U = -327000\ \text{cal} + (1)\,(2\ \text{cal mol}^{-1}\text{K}^{-1})\,(300\ \text{K})$$

$$\Delta U = -327000\ \text{cal} + 600\ \text{cal}$$

$$\Delta U = -326400\ \text{cal}$$

The negative sign denotes that heat is released. The magnitude of heat evolved at constant volume is therefore

$$326400\ \text{cal}$$

So, the answer is $$326400\ \text{cal}$$.