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# When $500J$ heat is given to a gas at constant pressure, it performs $\dfrac{{500}}{3}J$ work. Find the average number of degrees of freedom per molecule of gas.(A) $3$(B) $4$(C) $5$(D) $6$

Last updated date: 13th Jul 2024
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Hint:Here, the heat supplied to a gas at constant pressure and the work performed by the gas are given. You are asked to find the average number of degrees of freedom per molecule of gas. In order to solve this question, you need to consider the first law of thermodynamics which involves the heat exchange, the change in internal energy and also the work done. Write the internal energy in terms of the specific heat at constant volume and then write the specific heat at constant volume in terms of degree of freedom using the relation between the quantity $\gamma$ and degree of freedom.

The ratio of specific heat at constant pressure and specific heat at constant volume is given as $\gamma = \dfrac{{{C_P}}}{{{C_V}}}$. In terms of degree of freedom, this ratio is given as $\gamma = 1 + \dfrac{2}{f}$. We know that ${C_P} - {C_V} = R$, let us derive a relation between ${C_V}$ and $f$.
$\gamma = 1 + \dfrac{2}{f} \\ \Rightarrow\dfrac{{{C_P}}}{{{C_V}}} = 1 + \dfrac{2}{f} \\ \Rightarrow\dfrac{{R + {C_V}}}{{{C_V}}} = 1 + \dfrac{2}{f} \\ \Rightarrow\dfrac{R}{{{C_V}}} + 1 = 1 + \dfrac{2}{f} \\ \Rightarrow{C_V} = \dfrac{{fR}}{2} \\$
The first law of thermodynamics is as follows, the heat exchange is equal to the sum of internal energy and the work done. Mathematically, $\Delta Q = \Delta U + W$. The heat supplied to the gas and the work done by the gas are given to you. Now, internal energy at constant volume is given as $\Delta U = n{C_V}\Delta T$. So, finally we can $\Delta U = \dfrac{f}{2}nR\Delta T$.
Now, the work done at constant pressure is given by $W = nR\Delta T$ which is given to be $\dfrac{{500}}{3}J$. Hence, we have, $nR\Delta T = \dfrac{{500}}{3}$. Also, we are given that $\Delta Q = 500J$. So, let us substitute all we have in the first equation.
$\Delta Q = \Delta U + W \\ \Rightarrow 500 = \dfrac{f}{2}nR\Delta T + \dfrac{{500}}{3} \\ \Rightarrow 500 = \dfrac{f}{2}\left( {\dfrac{{500}}{3}} \right) + \dfrac{{500}}{3} \\ \Rightarrow 500 = \dfrac{{500}}{3}\left( {\dfrac{f}{2} + 1} \right) \\ \Rightarrow 3 = \dfrac{f}{2} + 1 \\ \therefore f = 4 \\$
Therefore, the average number of degrees of freedom per molecule of gas is equal to $4$.
Note:We have considered many things such as first law of thermodynamics, heat exchange, internal energy, work done, degree of freedom, specific heats and their ratio, you need to keep in and all these quantities also the formula for each so that you can express one quantity in terms of other. Note that here we have written internal energy in terms of specific heat at constant volume, it does not mean that we have changed the process from constant pressure to constant volume. The internal energy is a state function and not a path function, meaning that no matter what path or process you take, the final change in internal energy would be the same. Internal energy can always be written as $\Delta U = n{C_V}\Delta T$.