Question

When heat $Q$ is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by $\Delta T$ . The heat required to produce the same change in temperature, at a constant pressure is:
(A) $\dfrac{7}{5}Q$
(B) $\dfrac{3}{2}Q$
(C) $\dfrac{5}{3}Q$
(D) $\dfrac{2}{3}Q$

Answer 1

Hint The heat required to produce change in temperature at constant pressure, this is asked, so the expression will have molar specific heat at constant pressure. So, we can use the relation between molar specific heat at constant volume and pressure to find it.
Formula Used
\[Q = n{C_p}\Delta T\]
${C_P} - {C_V} = R$

Complete step by step answer
When a body is heated, the temperature is increased. Suppose the rise in temperature is $\Delta T$ and m is the mass, then the heat required Q to raise by this temperature will be
\[Q = mc\Delta T\]where c is the specific heat of the substance and Q is the amount of heat.
From the question, it is said that the amount of heat required by a diatomic gas to raise its temperature by $\Delta T$ at constant volume is given to be Q, so Q will be
\[Q = n{C_v}\Delta T\] where ${C_v}$ is the molar heat capacity at constant volume
A diatomic gas is given. So, for it we know that degree of freedom is 5. Degree of freedom means that a diatomic gas has 3 translational and 2 rotational motion.
And specific heat of gas at constant pressure in terms of degree of freedom is given as
${C_V} = \dfrac{1}{2}fR$
For diatomic gas, ${C_V} = \dfrac{5}{2}R$ as $f = 5$
The amount of heat required by a gas to raise its temperature $\Delta T$ at constant volume Q
$Q = n{C_V}\Delta T$, put the value of ${C_V}$ Then,
$ \Rightarrow Q = \dfrac{{5R}}{2}n\Delta T$
$ \Rightarrow nR\Delta T = \dfrac{2}{5}Q$----- (a)
At constant pressure, specific molar heat will be ${C_P}$ and we know,
${C_P} - {C_V} = R$
$ \Rightarrow {C_P} = {C_V} + R$
$ \Rightarrow {C_P} = \dfrac{{5R}}{2} + R = \dfrac{{7R}}{2}$
The amount of heat required by a gas to raise its temperature $\Delta T$ at constant volume Q’
$Q' = n{C_P}\Delta T$ put the value of ${C_P}$
$ \Rightarrow Q' = \dfrac{7}{2}(Rn\Delta T)$ putting values from equation (a)
 $ \Rightarrow Q' = \dfrac{7}{2} \times \dfrac{2}{5}Q = \dfrac{7}{5}Q$

Hence, the correct option is A.

Note
Degree of freedom is the number of independent motions by a particle. It can also be understood as the number of independent methods a particle exchanges energy. For diatomic gases, two atoms are joined to one another through a bond. So, they can not only move along three coordinate axes but also rotate on any of the axes. But the moment of inertia about axes is negligible so it has only two rotational axes. Hence the degree of freedom is 5.