# Question 9802d

Mar 5, 2016

${\text{0.59 J K}}^{- 1}$

#### Explanation:

Your starting point here will be the fundamental thermodynamic relation, which establishes a relationship between infinitesimal changes in internal energy, entropy, and volume for a closed system at thermal equilibrium

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \mathrm{dU} = T \mathrm{dS} - P \mathrm{dV} \textcolor{w h i t e}{\frac{a}{a}}}} |} \text{ }$, where

$U$ - the internal energy of the gas
$S$ - its entropy
$V$ - the volume it occupies
$P$ - the pressure of the gas
$T$ - the absolute temperature of the gas

Now, an isothermal expansion is characterized by the fact that the temperature of the gas remains unchanged. For an ideal gas that undergoes an isothermal expansion, the fact fact that temperature is held constant implies that you have

$\mathrm{dU} = 0$

This means that you can say

$0 = T \mathrm{dS} - P \mathrm{dV} \implies T \mathrm{dS} = P \mathrm{dV}$

This can be rearranged to give

$\mathrm{dS} = \frac{P}{T} \mathrm{dV}$

Now, assuming that we're going from an initial state $\textcolor{p u r p \le}{\left(1\right)}$ of ${V}_{1}$ and $T$ to a final state $\textcolor{p u r p \le}{\left(2\right)}$ of ${V}_{2}$ and $T$, you can find the change in entropy, $\Delta S$, by integrating the above equation

$\left[\mathrm{dS} = \frac{P}{T} \mathrm{dV}\right] \to {\int}_{\textcolor{p u r p \le}{1}}^{\textcolor{p u r p \le}{2}}$

$\Delta S = {\int}_{\textcolor{p u r p \le}{1}}^{\textcolor{p u r p \le}{2}} \frac{P}{T} \mathrm{dV}$

Since you're dealing with an ideal gas, you can use the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}}}} |}$

to say that

$\frac{P}{T} = \frac{n R}{V} = {\overbrace{n R}}^{\textcolor{red}{\text{constant}}} \cdot \frac{1}{V}$

This means that you have

$\Delta S = n R \cdot {\int}_{\textcolor{p u r p \le}{1}}^{\textcolor{p u r p \le}{2}} \frac{1}{V} \mathrm{dV}$

Finally, this will get you

$\Delta S = n R \cdot \ln \left({V}_{2} / {V}_{1}\right)$

Here

$n$ - the number of moles of gas present in the sample.
$R$ - the universal gas constant, equal to ${\text{8.314 J K"^(-1)"mol}}^{- 1}$

Plug in your values to get

DeltaS = 0.13 color(red)(cancel(color(black)("moles"))) * "8.314 J"color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1) * ln((19 color(red)(cancel(color(black)("L"))))/(11color(red)(cancel(color(black)("L")))))#

$\Delta S = \textcolor{g r e e n}{| \overline{\underline{{\text{0.59 J K}}^{- 1}}} |}$

Now, does a positive change in entropy make sense?

Notice that when ${V}_{2} > {V}_{1}$, $\Delta S > 0$. This tells you that increasing the volume of a gas while keeping its temperature constant will result in an increase in entropy.

That happens because the same number of molecules of gas are now floating around in a larger volume, which implies that the randomness and disorder of the system have increased.