What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{\text {ext }}$ in a single step as shown in figure? Explain graphically.
Suppose total volume of the gas is $V_i$ and pressure of the gas inside cylinder is $p$. After compression by constant external pressure, $\left(p_{\text {ext }}\right)$ in a single step, final volume of the gas becomes $V_f$.
Then volume change, $\Delta v=\left(V_f-V_i\right)$
If $W$ is the work done on the system by movement of the piston, then
$$\begin{aligned} & W=p_{\text {ext }}(-\Delta V) \\ & W=-p_{\text {ext }}\left(V_f-V_i\right) \end{aligned}$$
This can be calculated from $p-V$ graph as shown in the figure. Work done is equal to the shaded area $A B V_f V_i$
The negative sign in this expression is required to obtain conventional sign for $W$ which will be positive. Because in case of compression work is done on the system, so $\Delta V$ will be negative.
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 .
$$\begin{aligned} & W=-2.303 n R T \log \frac{p_1}{p_2} \\ & W=-2.303 \times 1 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{Klog} 2 \quad \left(\because \frac{p_1}{p_2}=\frac{2}{1}\right)\\ & W=-2.303 \times 1 \times 8.314 \times 298 \times 0.3010 \mathrm{~J} \\ & W=-1717.46 \mathrm{~J} \end{aligned}$$