Question

A cylinder containing an ideal gas $\left(0.1\mathrm{mol}\mathrm{of}1.0{\mathrm{dm}}^3\right)$ is in thermal equilibrium with a large volume of $0.5\mathrm{Molal}$ aqueous solution of ethylene glycol at its freezing point. If the stoppers ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ (as shown in the figure) suddenly withdrawn, the volume of the gas in $\mathrm{Liters}$ after equilibrium is achieved will be:

(Given, ${\mathrm{K}}_{\mathrm{f}}\left(\mathrm{water}\right)=2.0\mathrm{K}{\mathrm{kgmol}}^{-1},\mathrm{R}=0.08{\mathrm{dm}}^3\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$)

Answer 1

Answer: $2.176{\mathrm{dm}}^3$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}=2 \\ \mathrm{Molality}\left(\mathrm{m}\right)=0.5 \\ ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ =2\times 0.5 \\ =1 \end{gathered}$$

So, the initial temperature now becomes $272\mathrm{K}$. Further using the given value of moles and initial volume of the gas and the calculated initial temperature value, we can find out the initial pressure of the ideal gas contained inside the piston.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{gas}}=\mathrm{nRT}/{\mathrm{V}}_1 \\ =\frac{(0.1)(0.08)\left(272\right)}{1} \\ =2.176\mathrm{atm} \end{gathered}$$

Now, on releasing the piston against an external pressure of $1\mathrm{atm}$, the gas will expand until the final pressure of the gas, i.e. ${\mathrm{P}}_2$ becomes equal to $1\mathrm{atm}$. During this expansion, since no reaction is happening and the temperature of the gas is not changing as well, Boyle’s law relation can be applied.

$$\begin{gathered} {\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{P}}_2{\mathrm{V}}_2 \\ 2.176\times 1=1\times {\mathrm{V}}_2 \\ {\mathrm{V}}_2=2.716{\mathrm{dm}}^3 \end{gathered}$$