Question

Find the change in entropy (in cal/k) of ${\text{1}}$ mole of ${{\text{O}}_{\text{2}}}$ gas $\left( {{{\text{C}}_{\text{v}}} = \dfrac{5}{2}{\text{R}}} \right)$, when it is
i.Heated from ${\text{300K}}$ to ${\text{400K}}$ isobarically
ii.Heated from ${\text{300K}}$ to ${\text{400K}}$ isochorically (Given $\ln 3 = 1 \cdot 1,\ln 2 = 0 \cdot 7$)

Answer 1

Hint:Entropy can be defined as the measure of randomness of the system. Entropy change is an extensive property. The isobaric process can be defined as a process during which the pressure of the system remains constant while the isochoric process is a process during which the volume of the system remains constant.

Complete step by step answer:
1) First of all we will learn to calculate the change in entropy in the isobaric process by using the formula given below:
$\Delta {\text{S}} = {\text{n}}{{\text{C}}_{\text{p}}}{\text{ln}}\left( {\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}} \right)$ $......\left( 1 \right)$
Where,
${{\text{T}}_1}$ = Initial temperature i.e. 300 K
${{\text{T}}_{\text{2}}}$ = Final temperature i.e. 400 K
n = 1 mole
2) The value of ${{\text{C}}_{\text{p}}}$ can be calculated from the given value of ${{\text{C}}_v}$ as:
${{\text{C}}_{\text{p}}} - {{\text{C}}_{\text{v}}} = {\text{nR}}$
Now, lets put the value of ${{\text{C}}_v}$ in the above formula we get,
${{\text{C}}_{\text{p}}} - \dfrac{5}{2}{\text{R}} = 1 \times {\text{R}}$
By taking the ${{\text{C}}_{\text{p}}}$ value on one side we get,
${{\text{C}}_{\text{p}}} = {\text{R}} + \dfrac{{\text{5}}}{{\text{2}}}{\text{R}}$
By doing the addition part in the above equation we get,
${C_P} = \dfrac{7}{2}{\text{R}}$
3) Now, let's substitute this value of ${{\text{C}}_{\text{p}}}$ in equation (1), we get
$\Delta {\text{S}} = {\text{1}} \times \dfrac{7}{2}{\text{R}} \times {\text{ln}}\left( {\dfrac{{{\text{400}}}}{{{\text{300}}}}} \right)$
By putting the values of R and logarithm we get,
$\Delta {\text{S}} = {\text{1}} \times \dfrac{7}{2} \times 8 \cdot 314 \times {\text{ln}}\left( {1 \cdot 33} \right)$
By putting the values of logarithm we get,
$\Delta {\text{S}} = {\text{1}} \times \dfrac{7}{2} \times 8 \cdot 314 \times 0 \cdot 2851$
By doing the above calculation we get,
$\Delta S = 2 \cdot 1\;{\text{cal}}{{\text{T}}^{ - 1}}$
4) Now the change in entropy in isochoric process can be calculated using the formula given below:
$\Delta {\text{S}} = {\text{n}}{{\text{C}}_{\text{v}}}{\text{ln}}\left( {\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}} \right)$ $......\left( 2 \right)$
Where,
${{\text{T}}_1}$ = Initial temperature i.e. 300 K
${{\text{T}}_{\text{2}}}$ = Final temperature i.e. 400 K
n = 1 mole
${{\text{C}}_{\text{v}}} = \dfrac{5}{2}{\text{R}}$
Now, let us substitute the value in equation (2), we get
$\Delta {\text{S}} = {\text{1}} \times \dfrac{5}{2}{\text{R}} \times {\text{ln}}\left( {\dfrac{{{\text{400}}}}{{{\text{300}}}}} \right)$
By putting the values of R and logarithm we get,
$\Delta {\text{S}} = {\text{1}} \times \dfrac{5}{2} \times 8 \cdot 314 \times {\text{ln}}\left( {1 \cdot 33} \right)$
By putting the values of logarithm we get,
$\Delta {\text{S}} = {\text{1}} \times \dfrac{5}{2} \times 8 \cdot 314 \times 0 \cdot 2851$
By doing the above calculation we get,
$\Delta S = 1 \cdot 5\;{\text{cal}}{{\text{T}}^{ - 1}}$
5) Therefore we got the answers as,
i) $\Delta S = 2 \cdot 1\;{\text{cal}}{{\text{T}}^{ - 1}}$
ii) $\Delta S = 1 \cdot 5\;{\text{cal}}{{\text{T}}^{ - 1}}$

Note:
Entropy (S) is a state function that does not depend on the path followed. Therefore, entropy change depends on the initial and the final states only. Entropy change is an extensive property which means the entropy is dependent on the mass.