Question

A lonely party balloon with a volume of 2.40 L and containing 0.100 mol of air is left behind to drift in the temporarily uninhabited and depressurized International Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is $\mathbf{425}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$. Calculate the entropy change of the air during the expansion.

Short answer

The change in entropy of air during the expansion is $10.0\mathrm{J}/\mathrm{K}$ .

Consider the given data as below.

Initial volume of air ${\mathrm{V}}_1=2.40\mathrm{L}=2.4\times {10}^{-3}{\mathrm{m}}^3$

Number of moles $\mathrm{n}=0.10\mathrm{mol}$

Final volume of air ${\mathrm{V}}_2=425{\mathrm{m}}^3$

Step-by-step solution

Step 1:  Definition of entropy

Entropy can be defined as the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. It is also the measure of randomness of system.

Formula of change in entropy is as below.

$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$

Where, $∆\mathrm{S}$is the change in entropy, $\mathrm{Q}$is the heat and $\mathrm{T}$is the temperature.

From first law of thermodynamics

$∆\mathrm{S}=\frac{\mathrm{W}+∆\mathrm{U}}{\mathrm{T}}$

Where, $∆\mathrm{S}$is the change in entropy of gas, $\mathrm{T}$is the temperature of gas, $\mathrm{W}$ is the work done by system and$∆\mathrm{U}$is change in internal energy.

Defining free expansion of gas

Such a expansion of a gas in which gas expands in such a way that no heat enters or leaves the system and also no work is done by or on the system. This type of expansion is known as free expansion of gas.

In free expansion both adiabatic and isothermal processes take place together.

Calculation of change in entropy of air

Consider the given data as below.

Initial volume of air,${\mathrm{V}}_{\mathrm{i}}=2.40\mathrm{L}=2.4\times {10}^{-3}{\mathrm{m}}^3$

Number of moles,$\mathrm{n}=0.10\mathrm{mol}$

Final volume of air, ${\mathrm{V}}_{\mathrm{f}}=425{\mathrm{m}}^3$

Using formula

$∆\mathrm{S}=\frac{\mathrm{W}+∆\mathrm{U}}{\mathrm{T}}$

In case of free expansion $∆\mathrm{U}=0$_.Therefore,

$\mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{\mathrm{Vi}}\right)$

Where, $\mathrm{W}$ is the work done by the system, n the number of moles, T is the temperature, ${\mathrm{V}}_{\mathrm{i}}$is the initial volume, ${\mathrm{V}}_{\mathrm{f}}$is the final volume, and R is the universal gas constant which is equal to $8.314\mathrm{J}/\mathrm{K}$_.

Therefore, the change in entropy will be,

$∆S=\frac{\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{\mathrm{Vi}}\right)}{\mathrm{T}}$

Now, putting the values of constants in above equation

$$\begin{gathered} ∆\mathrm{S}=0.10\times 8.314\times \mathrm{In}\left(\frac{425}{2.4\times {10}^{-3}}\right) \\ =0.10\times 8.314\times \mathrm{In}\frac{425}{2.4\times {10}^{-3}} \\ =10.0\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the change in entropy of air during the expansion is $10.0\mathrm{J}/\mathrm{K}$_.