The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression $W=-n R T \ln \frac{V_f}{V_i}$. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below
$$2 \mathrm{Zn}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{ZnO}(s) ; \Delta H=-693.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
18.0 g of water completely vaporises at $100^{\circ} \mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is $40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?
Given that, quantity of water $=18.0 \mathrm{~g}$, pressure $=1$ bar
As we know that, $18.0 \mathrm{~g~H}_2 \mathrm{O}=1 \mathrm{~mole~H}_2 \mathrm{O}$
Enthalpy change for vaporising $1 \mathrm{~mole}^{-1} \mathrm{H}_2 \mathrm{O}=40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ Enthalpy change for vaporising 2 moles of $\mathrm{H}_2 \mathrm{O}=2 \times 40.79 \mathrm{~kJ}=81.358 \mathrm{~kJ}$
Standard enthalpy of vaporisation at $100^{\circ} \mathrm{C}$ and 1 bar pressure, $\Delta_{\text {vap }} H^{\circ}=+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
One mole of acetone requires less heat to vaporise than 1 mole of water. Hence, acetone has less enthalpy of vaporisation and water has higher enthalpy of vaporisation. It can be represented as $\left(\Delta H_V\right)$ water $>\left(\Delta H_V\right)$ acetone.