Question

To raise the temperature of a certain mass of gas by \[{50^ \circ }C\] at constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by \[{100^ \circ }C\]at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

Answer 1

Hint: In this question, we need to find the number of degrees of freedom that each molecule of the gas has. For this, we will use the concept of molar heat capacity to get the desired result.

Formula used: The formula for molar heat capacity is given below.
\[q = n{C_p}\vartriangle t\]
Here, \[q\] is the heat energy, \[n\] is the number of moles of a gas, \[{C_p}\] is the constant pressure and \[\vartriangle t\] is the change in temperature.
Similarly, we can say that for constant volume process, the heat energy is given by
\[q = n{C_V}\vartriangle t\]
Here, \[{C_V}\] is the constant volume.
Also, degrees of freedom of a gas is given by
\[f = \dfrac{2}{{\gamma - 1}}\] where, \[\gamma = \dfrac{{{C_p}}}{{{C_V}}}\]

Complete step by step solution:
The molar heat capacity for constant pressure process is given by
\[q = n{C_p}\vartriangle t\]
Here, \[q = 160\] calories, \[\vartriangle t = 50\]
Thus, we get
\[160 = n{C_p}\left( {50} \right)\]….. (1)
The molar heat capacity for constant volume process is given by
\[q = n{C_V}\vartriangle t\]
Here, \[q = 240\] calories, \[\vartriangle t = 100\]
Thus, we get
\[240 = n{C_V}\left( {100} \right)\]….. (2)

By taking the ratio of (1) and (2), we get
\[\dfrac{{160}}{{240}} = \dfrac{{n{C_p}}}{{n{C_V}}} \times \dfrac{{50}}{{100}}\]
By simplifying, we get
\[\dfrac{{16}}{{24}} = \dfrac{{{C_p}}}{{2{C_V}}}\]
\[\Rightarrow \dfrac{{16}}{{24}} \times 2 = \dfrac{{{C_p}}}{{{C_V}}}\]
\[\Rightarrow \dfrac{4}{3} = \dfrac{{{C_p}}}{{{C_V}}}\]
But we can say that \[\gamma = \dfrac{{{C_p}}}{{{C_V}}}\]
So, degrees of freedom of a gas is given by
\[f = \dfrac{2}{{\gamma - 1}}\]
Now, put \[\dfrac{4}{3} = \gamma \] in the above equation, we get
\[f = \dfrac{2}{{\dfrac{4}{3} - 1}}\]

By simplifying, we get
\[f = \dfrac{2}{{\dfrac{{4 - 3}}{3}}}\]
\[\Rightarrow f = \dfrac{2}{{\dfrac{1}{3}}}\]
\[\Rightarrow f = 2 \times 3\]
\[\therefore f = 6\]
Hence, the amount of degrees of freedom that each molecule of the gas has is 6.

Therefore, there are 6 degrees of freedom that each molecule of the gas has.

Note:Many students generally make mistakes in writing the formula of molar heat capacity for constant pressure and for constant volume because of which they may get wrong in calculating \[\dfrac{{{C_p}}}{{{C_V}}}\].