Question

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increased the internal energy of the gas is:
(A). $\dfrac{2}{5}$
(B). $\dfrac{3}{5}$
(C). $\dfrac{3}{7}$
(D). $\dfrac{5}{7}$

Answer 1

Hint: These types of question follows the basic patterns; you just have to remind some basic formulas for example ${C_v} = \dfrac{f}{2}$ and ${C_p} = \dfrac{f}{2} + 1$ also the formula $\dfrac{{{C_v}}}{{{C_p}}} = \dfrac{1}{\gamma }$ to find the correct option.

Complete step-by-step solution -

Total heat energy supplied to raise a diatomic gas's temperature at constant pressure = $n{C_p}\vartriangle T$
And the total rise in internal energy = $n{C_v}\vartriangle T$
Thus, fraction of heat energy utilized in increasing internal energy = $\dfrac{{n{C_v}\vartriangle T}}{{n{C_p}\vartriangle T}}$ (equation 1)
Simplifying equation 1
\Rightarrow $$$\dfrac{{{C_v}}}{{{C_p}}} = \dfrac{{\dfrac{f}{2}}}{{\dfrac{f}{2} + 1}}$$ Since we know that $${C_v} = \dfrac{f}{2}$$ and $${C_p} = \dfrac{f}{2} + 1$$ \Rightarrow \dfrac{{{C_v}}}{{{C_p}}} = \dfrac{f}{{f + 2}}$$ (Equation 2) Here f is the number of degrees of freedom of the gas and here for diatomic gas f = 5. Substituting the value of f in the equation 2 $ \Rightarrow \dfrac{{{C_v}}}{{{C_p}}} = \dfrac{5}{{5 + 2}}$Since$\dfrac{{{C_v}}}{{{C_p}}} = \dfrac{1}{\gamma }$Therefore$\dfrac{1}{\gamma } = \dfrac{5}{7}$$ is a fraction of the heat energy supplied which increases the internal energy of the gas.
Hence, D is the correct option.

Note: We have noticed that the term internal energy of an ideal gas is the main concept in this question and it can be explained as the internal changes in energy in an ideal gas can be represented only by changes in its kinetic energy. Kinetic energy is simply the perfect gas’s internal energy and depends entirely on its pressure, volume, and thermodynamic temperature. An ideal gas's intrinsic energy is proportional to its mass (number of moles) n and its temperature T.i.e. $U = cnT$