Find the number of degrees of freedom of molecules in a gas. Whose molar heat capacity at constant pressure \[{C_{\text{P}}} = 29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - 1}}\].
Answer
588.9k+ views
Hint:In this question use the concept of the adiabatic index, that is it is the ratio of the specific heat at constant pressure and the specific heat at constant volume. First, calculate the specific heat at constant volume and then calculate the adiabatic index.
Complete step by step answer:
As we know that the molar heat capacity of any substance is the amount of energy that is in the form of heat added to the substance to one mole of the substance because of this the temperature of the substance increases.
We know that the constant pressure means that the heat transfer to the system does work but it changes the internal energy of the system. We are given molar heat capacity at constant pressure \[{C_{\text{P}}} = 29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}\,{{\text{K}}^{ - 1}}\]. First we calculate the value of \[{C_{\text{v}}}\], use the ideal gas equation formula \[{\text{R}}\] is the gas constant and the value of \[{\text{R}}\] is \[8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}\].
\[{C_{\text{p}}} - {C_{\text{v}}} = {\text{R}}\]
\[ \Rightarrow {C_{\text{v}}} = {C_{\text{p}}} - {\text{R}}\]
Now we substitute the given in the above equation,
\[{C_{\text{v}}} = 29 - 8.314\]
\[ \Rightarrow {C_{\text{v}}} = 20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}\]
Now, use the formula to calculate the adiabatic index $\gamma $.
\[\gamma = \dfrac{{{C_{\text{P}}}}}{{{C_{\text{V}}}}}\]
Now we substitute the values in the above equation,
\[\gamma = \dfrac{{29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}{{20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}\]
By simplification we get,
\[ \Rightarrow \gamma = 1.4\]
Now, we calculate the degrees of freedom of the molecules $\left( {\text{f}} \right)$ in a gas,
\[\sqrt {\dfrac{3}{\gamma }} = \sqrt {\dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}} \]
After squaring both sides, we get
\[ \Rightarrow \dfrac{3}{\gamma } = \dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}\]
By simplification we get,
\[ \Rightarrow {\text{f}} + 2 = \gamma {\text{f}}\]
Now, Substitute the value of $\gamma $ in the above equation,
\[{\text{f}} + 2 = 1.4{\text{f}}\]
\[\therefore {\text{f}} = 5\]
Therefore, the number of degrees of freedom of molecules in a gas is \[5\].
Note:
Do not confuse the universal gas constant and the characteristic gas constant. As we know that the value of universal gas constant is \[8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}\] and it is same for each gas, but characteristic gas constant depends on the molecular mass of the gas.
Complete step by step answer:
As we know that the molar heat capacity of any substance is the amount of energy that is in the form of heat added to the substance to one mole of the substance because of this the temperature of the substance increases.
We know that the constant pressure means that the heat transfer to the system does work but it changes the internal energy of the system. We are given molar heat capacity at constant pressure \[{C_{\text{P}}} = 29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}\,{{\text{K}}^{ - 1}}\]. First we calculate the value of \[{C_{\text{v}}}\], use the ideal gas equation formula \[{\text{R}}\] is the gas constant and the value of \[{\text{R}}\] is \[8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}\].
\[{C_{\text{p}}} - {C_{\text{v}}} = {\text{R}}\]
\[ \Rightarrow {C_{\text{v}}} = {C_{\text{p}}} - {\text{R}}\]
Now we substitute the given in the above equation,
\[{C_{\text{v}}} = 29 - 8.314\]
\[ \Rightarrow {C_{\text{v}}} = 20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}\]
Now, use the formula to calculate the adiabatic index $\gamma $.
\[\gamma = \dfrac{{{C_{\text{P}}}}}{{{C_{\text{V}}}}}\]
Now we substitute the values in the above equation,
\[\gamma = \dfrac{{29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}{{20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}\]
By simplification we get,
\[ \Rightarrow \gamma = 1.4\]
Now, we calculate the degrees of freedom of the molecules $\left( {\text{f}} \right)$ in a gas,
\[\sqrt {\dfrac{3}{\gamma }} = \sqrt {\dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}} \]
After squaring both sides, we get
\[ \Rightarrow \dfrac{3}{\gamma } = \dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}\]
By simplification we get,
\[ \Rightarrow {\text{f}} + 2 = \gamma {\text{f}}\]
Now, Substitute the value of $\gamma $ in the above equation,
\[{\text{f}} + 2 = 1.4{\text{f}}\]
\[\therefore {\text{f}} = 5\]
Therefore, the number of degrees of freedom of molecules in a gas is \[5\].
Note:
Do not confuse the universal gas constant and the characteristic gas constant. As we know that the value of universal gas constant is \[8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}\] and it is same for each gas, but characteristic gas constant depends on the molecular mass of the gas.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

