Question

An ideal diatomic gas is heated at a constant pressure. What fraction of heat energy is utilized to increase its kinetic energy?
$A.\;\;\dfrac{5}{7}$
$B.\;\;\dfrac{2}{5}$
$C.\;\;\dfrac{3}{5}$
$D.\;\;\dfrac{3}{7}$

Answer 1

Hint: We can use the relation between internal energy and temperature for an ideal gas to relate change in temperature and change in internal energy. We can then use the expression for heat capacity at constant pressure for an ideal gas obtained from kinetic interpretations.

Complete step-by-step answer:
We know that the internal energy of a gas molecule depends on the temperature of the gas and the degree of freedom of the molecules. A diatomic gas has 5 degrees of freedom. Thus, we can use the law of Equi-partition of energy to find the energy associated with each molecule.
For a given temperature $T$, the energy of one gas molecule would be :
$E' = 5 \times \dfrac{1}{2}{K_B}T$
Hence the total energy of one mole of the gas, which contains ${N_A}$ number of particles would be :
$E' = {N_A}\dfrac{5}{2}{K_B}T$ Where ${N_A}$ is the Avogadro number.
We know that ${K_B} = \dfrac{R}{{{N_A}}}$ where R is the gas constant. Thus, the total energy of the gas, which is the internal energy $U$ is
$U = {N_A}\dfrac{5}{2}\dfrac{R}{{{N_A}}}T = \dfrac{5}{2}R\;T$ (1)
Also, we know the first law of thermodynamics, which says :
$\Delta Q = \Delta U + P\Delta V$ which gives a relation between heat supplied $\Delta Q$ and the change in internal energy $\Delta U$. If the heat is supplied keeping the volume constant, The supplied heat would be equal to the change in internal energy.
$\Delta U = {\left. {\Delta Q} \right|_{V = const}}$
 Since we know $U$ from (1), we can use this to find the amount of heat supplied at constant volume to raise the temperature of the gas.
 ${\left. {\Delta Q} \right|_{V = const}} = \Delta U = \dfrac{5}{2}R\Delta T$
 So the heat capacity at constant volume for the gas is :
 ${C_v} = \dfrac{{\Delta V}}{{\Delta T}} = \dfrac{5}{2}R$

Now that we know${C_v}$, we can find ${C_p}$ as ${C_p} = {C_v} + R$
${C_p} = \dfrac{5}{2}R + R = \dfrac{7}{2}R$
Now let us say we gave $\Delta Q$ amount of heat to an Ideal gas at constant pressure. If the temperature rise due to this supplied heat is $\Delta T$, then we can say that
$\Delta Q = {C_p}\Delta T = \dfrac{7}{2}R\Delta T$
Also, the change in internal energy of the gas due to supplied heat would be
$\Delta U = \dfrac{5}{2}R\Delta T$
Thus the Ratio: $\dfrac{{\Delta U}}{{\Delta Q}} = \dfrac{5}{7}$
This means that $\dfrac{5}{7}$ of the heat supplied goes into changing the internal energy of the system. Hence the correct answer is option A.

Note: Note that the $P\Delta V$ term in $\Delta Q = \Delta U + P\Delta V$ indicates the work that the system does when given a heat energy of $\Delta Q$. Here we saw that $\dfrac{5}{7}$ of the heat supplied goes into changing the internal energy. Thus, only $\dfrac{2}{7}$ of the heat supplied is available as mechanical work.