How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
When compression is carried out in infinite steps with change in pressure, it is a reversible process. The work done can be calculated from $p-V$ plot as shown in the given figure. Shaded area under the curve represents the work done on the gas.
Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b) $\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HCl}(\mathrm{g}) \Delta_r H^{\mathrm{s}}=-92.32 \mathrm{~kJ} \mathrm{~mol}^{-1}$
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Representation of potential energy/enthalpy change in the following processes
(a) Throwing a stone from the ground to roof.
(b) $\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{Cl}_2(g) \rightleftharpoons \mathrm{HCl}(g) ; \Delta_r H^s=-92.32 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Energy increases in (a) and it decreases in (b) process. Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.
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.