Question

In the case of a diatomic gas, the ratio of the heat used in doing work for expansion of the gas to the total heat given to it at constant pressure is:
(A) $\dfrac{2}{5}$
(B) $\dfrac{3}{7}$
(C) $\dfrac{2}{7}$
(D) $\dfrac{5}{7}$

Answer 1

Hint : We know that the diatomic molecules are molecules that are composed of only two atoms, of the same or different chemical elements. Recall the expressions for change in heat, change in internal energy and work done in the constant pressure process.

Complete Step By Step Answer:
We know that the process where the pressure remains the same is known as isobaric process. For the Isobaric process, Change in heat energy, $\Delta Q=n{{C}_{P}}\Delta T$ where ${{C}_{P}}$ is molar heat capacity at a constant pressure. Change in the internal energy, $\Delta U=n{{C}_{V}}\Delta T$ where ${{C}_{V}}$ is molar heat capacity at a constant pressure. As the body absorbs heat the temperature of the body rises, but when heat is withdrawn from the body it cools down, so the body heat decreases. The temperature of any body is the measure of its molecules’ kinetic energy. Heat capacity is the ratio of heat absorbed by a material to the temperature change. Therefore, the temperature change in a body is directly proportional to the heat transferred to the given body.
In a diatomic gas we have ${{C}_{P}}=\dfrac{7}{2}R$ and ${{C}_{V}}=\dfrac{5}{2}R.$ The heat is given as $n{{C}_{P}}\Delta T$ and $n{{C}_{V}}\Delta T.$
Thus, we get is $\dfrac{U}{Q}=\dfrac{\left( \dfrac{5}{2}R \right)}{\left( \dfrac{7}{2}R \right)}=\dfrac{5}{2}\times \dfrac{2}{7}=\dfrac{5}{7}.$
$\Rightarrow \dfrac{U}{Q}=\dfrac{5}{7}.$
Therefore, the correct answer is option D.

Note :
Remember 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.