Answer
Verified
409.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE