Question

The molar entropy of helium gas at 25°C and 1.00 atm is $\mathbf{126}\mathbf{.}\mathbf{1}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}$. Assuming ideal behavior, calculate the entropy of the following.

1. 0.100 mole of $\mathit{H}\mathit{e}\mathbf{\left(}\mathit{g}\mathbf{\right)}$at 25°C and a volume of 5.00 L.
2. 3.00 moles of $\mathit{H}\mathit{e}\mathbf{\left(}\mathit{g}\mathbf{\right)}$at 25°C and a volume of 3000.0 L.
   

Short answer

The entropyof the following,

1. 0.100 mole of $He\left(g\right)$at 25°C and a volume of 5.00 L is$\mathbf{13}\mathbf{.}\mathbf{2}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}$

2. 3.00 moles of $He\left(g\right)$at 25°C and a volume of 3000.0 L is$\mathbf{471}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}$

Step-by-step solution

Step 1: Introduction to the Concept

The entropy change expression is calculated as follows,

$\mathit{\Delta }\mathit{S}\mathbf{=}\mathit{n}\mathit{R}\mathit{l}\mathit{n}\left(\frac{V_2}{V_1}\right)$

Where,

$\Delta S=$Entropy change

$n=$Number of moles

$R=$Universal gas constant

$V_2=$Final Volume

$V_1=$Initial volume

Determination of Change in entropy at 0.100 mole of He(g) at 25°C and a volume of 5.00 L

a)

Initial entropy is computed,

$S_i=0.100\text{mol}\times \text{126.1J/Kmol}$

$=12.6J/K$

Initial volume is given as,

$V_1=\frac{0.100mol\times \text{0.08206Latm/Kmol}\times \text{29.8 K}}{1.00\text{atm}}$

$V_1=2.45\text{L}$

Change in entropy is given as,

$S_f-S_i=0.100\text{mol}\times \text{8.314J/Kmol}\times \ln \left(\frac{5.00\text{L}}{2.45\text{L}}\right)$

$S_f=12.6+0.593$

$S_f=13.2J/K$

Therefore the entropy change is $\mathbf{13}\mathbf{.}\mathbf{2}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}$.

Determination of Change in entropy at 0.100 mole of He(g)at 25°C and a volume of 5.00 L

b)

Initial entropyis computed,

$S_i=3.00\text{mol}\times \text{126.1J/Kmol}$

$=378\text{J/K}$

Initial volume is given as,

$V_1=\frac{3.00mol\times 0.08206Latm/Kmol\times \text{29.8 K}}{1.00\text{atm}}$

$V_1=73.4\text{L}$

Change in entropy is given as,

$S_f-S_i=3.00\text{mol}\times \text{8.314J/Kmol}\times \ln \left(\frac{3000.0\text{L}}{73.4\text{L}}\right)$

$S_f=378+92.6$

role="math" localid="1658763733490" $S_f=471J/K$

Therefore the entropy change is$\mathbf{471}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}$.