Enthalpy diagram for a particular reaction is given in figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain.
No, enthalpy is one of the contributing factors in deciding spontaneity but it is not the only factor. Another contributory factor, entropy factor has also to be taken into consideration.
1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K .
The given diagram represent that the process is carried out in infinite steps, hence it is isothermal reversible expansion of the ideal gas from pressure 2.0 atm to 1.0 atm 298 K .
$$\begin{aligned} & W=-2.303 n R T \log \frac{p_1}{p_2} \\ & W=-2.303 \times 1 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{Klog} 2 \quad \left(\because \frac{p_1}{p_2}=\frac{2}{1}\right)\\ & W=-2.303 \times 1 \times 8.314 \times 298 \times 0.3010 \mathrm{~J} \\ & W=-1717.46 \mathrm{~J} \end{aligned}$$
An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that, 1 L bar = 100 J )
In the first case, as the expansion is against constant external pressure
$$\begin{aligned} W & =-p_{\text {ext }}\left(V_2-V_1\right)=-2 \text { bar } \times(50-10) \mathrm{L} \\ & =-80 \mathrm{~L} \text { bar } \quad \text{(1L bar = 100 J)}\\ & =-80 \times 100 \mathrm{~J} \\ & =-8 \mathrm{~kJ} \end{aligned}$$
If the given expansion was carried out reversibly, the internal pressure of the gas should be greater than the external pressure at every stage. Hence, the work done will be more.
Match the following.
| List I | List II | ||
|---|---|---|---|
| A. | Adiabatic process | 1. | Heat |
| B. | Isolated system | 2. | At constant volume |
| C. | Isothermal change | 3. | First law of thermodynamics |
| D. | Path function | 4. | No exchange of energy and matter |
| E. | State function | 5. | No transfer of heat |
| F. | $$\Delta U=q$$ | 6. | Constant temperature |
| G. | Law of conservation of energy | 7. | Internal energy |
| H. | Reversible process | 8. | $$p_{ext=0}$$ |
| I. | Free expansion | 9. | At constant pressure |
| J. | $$\Delta H=q$$ | 10. | Infinitely slow process which proceeds through a series of equilibrium states. |
| K. | Intensive property | 11. | Entropy |
| L. | Extensive property | 12. | Pressure |
| 13. | Specific heat |
$$\begin{aligned} &\begin{aligned} & \text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}\\ &\text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}$$
Correct matching can be done as
| A. | Adiabatic process | No transfer of heat |
|---|---|---|
| B. | Isolated system | No exchange of energy and matter |
| C. | Isothermal change | Constant temperature |
| D. | Path function | Heat |
| E. | State function | Internal energy Entropy Pressure |
| F. | $$\Delta U=q$$ | At constant volume |
| G. | Law of conservation of energy | First law of thermodynamics |
| H. | Reversible process | Infinitely slow process which proceeds through a series of equilibrium states |
| I. | Free expansion | $$p_{ext}=0$$ |
| J. | $$\Delta H=q$$ | At constant pressure |
| K. | Intensive property | Heat Pressure Specific heat |
| L. | Extensive property | Internal energy Entropy |
Match the following processes with entropy change
| Reaction | Entropy change | ||
|---|---|---|---|
| A. | A liquid vaporises | 1. | $$\Delta S=0$$ |
| B. | Reaction is non-spontaneous at all | 2. | $$\Delta S$$ = positive |
| C. | Reversible expansion of an ideal gas | 3. | $$\Delta S$$ = negative |
$$\mathrm{A.\to(2)\quad B.\to(3)\quad C.\to(1)}$$
A. When a liquid vaporises, i.e., liquid $\rightarrow$ vapour, entropy increase i.e., $\Delta S=$ positive
B. When $\Delta H=$ positive, i.e., energy factor opposes. The process is non-spontaneous at all temperatures if entropy factor also opposes the process, i.e., $\Delta S=$ negative
C. In the reversible expansion of an ideal gas, the system remains in equilibrium at every stage. Hence, $\Delta S=0$