91Ó°ÊÓ

The heat transferred to the monatomic gas Is Q if the number of moles is n=1mol.

Using the relation between specific heat and internal energy of monatomic and diatomic gas, we can find the relation between monatomic gas volume and diatomic gas volume at a constant pressure. After that, we can find the relation between the monatomic and diatomic gas temperatures at a constant volume; from that, we can answer the above questions.

Formulae:

Work done by the gas at constant press pv=nRT sure, $W=\underset{v_i}{\overset{v_f}{}}pdv$ …(¾±)

Change in internal energy at constant volume,$\mathrm{Δ}{E}_{int}=n{C}_v\mathrm{Δ}T$ …(¾±¾±)

Ideal gas equation for n number of mol,…(iii)

According to first law of thermodynamics, the change in internal energy is given by

$\mathrm{Δ}E=Q-W$ …(¾±±¹)

$\text{Δ·¡}={\text{nC}}_v\text{Δ°Õ}$

For monatomic gas:

The value of specific heat at constant volume is $C_v=\frac{3}{2}R$

The value of specific heat at constant pressure is $C_p=\frac{5}{2}R$

For diatomic gas:

The value of specific heat at constant volume is $C_v=\frac{5}{2}R$.

The value of specific heat at constant pressure is $C_p=\frac{7}{2}R$.

Change in internal energy is and work done is, $W=p\mathrm{Δ}V.$

Substituting the values of work from equation (i) and change in internal energy from equation (ii) in equation (iv), we get

$$\begin{gathered} n{C}_v\mathrm{Δ}T=Q-p\mathrm{Δ}V \\ Q=n{C}_v\mathrm{Δ}T+p\mathrm{Δ}V \end{gathered}$$ …(±¹)

Now, using equation (iii), we get the value of change in temperature as follows:

$\mathrm{Δ}T=\frac{p\mathrm{Δ}V}{nR}$

Putting this value in equation (v), we get the equation of heat as

$$\begin{gathered} Q=n{C}_v\frac{p\mathrm{Δ}V}{nR}+p\mathrm{Δ}V \\ =\left(\frac{C_v}{R}+1\right)p\mathrm{Δ}V \end{gathered}$$ …(±¹¾±)

For monatomic gas, using the given values in equation (vi), we can get the heat as follows:

$$\begin{gathered} Q=\left(\frac{\frac{3}{2}R}{R}+1\right)p\mathrm{Δ}{V}_m(\text{where subscript m is for monoatomic gas)} \\ =\frac{5}{2}p\mathrm{Δ}{V}_m \end{gathered}$$ …(±¹¾±¾±)

Similarly, using the given values in equation (vi), we can get the heat for diatomic gas as follows:

$$\begin{gathered} Q=\left(\frac{\frac{5}{2}R}{R}+1\right)p\mathrm{Δ}{V}_d(\text{where subscript d is for diatomic gas)} \\ =\frac{7}{2}p\mathrm{Δ}{V}_d \end{gathered}$$ …(±¹¾±¾±¾±)

Now, from equations (vii) and (viii), we get the volume relation as

$$\begin{gathered} \frac{5}{2}p\mathrm{Δ}{V}_m=\frac{7}{2}p\mathrm{Δ}{V}_d \\ 5V_m=7V_d \end{gathered}$$

That means the volume of monatomic gas is greater than the volume of diatomic gas$\mathrm{Δ}{V}_m>\mathrm{Δ}{V}_d$

For any constant volume process

$W=0$

Thus, the value of heat is given by equation (v) as follows: $Q=n{C}_v\mathrm{Δ}T$.

For monatomic gas, the value of heat is given using the value of specific heat by

$Q=n\frac{3}{2}R\mathrm{Δ}T$ …(¾±³æ)

For diatomic gas, the value of heat is given using the value of specific heat by

$Q=n\frac{5}{2}R\mathrm{Δ}T$ …(³æ)

Now equating (ix) and (x), we can write the temperature relation as follows:

$\begin{array}{c}n\frac{3}{2}R\mathrm{Δ}{T}_m=n\frac{5}{2}R\mathrm{Δ}{T}_d\\ 3\mathrm{Δ}{T}_m=5\mathrm{Δ}{T}_d\end{array}$

i.e.,$T_m>T_D$

From the graph, we can conclude that for a mono atomic gas at constant pressure, path 3 is related with a monatomic gas. From condition $\mathrm{Δ}{V}_m>\mathrm{Δ}{V}_d$, we can say that path 3 is related with an isobaric process.

From the graph, we can conclude that for a mono atomic gas at constant volume, path 1 is related with a mono atomic gas. From condition, $T_m>T_D$, we can say that is related with an isochoric process.

From the graph, we can conclude that for a diatomic gas at constant pressure, is related with a diatomic gas. From condition, $V_m>V_d$, we can say that is related with an isobaric process.

From the graph, we can conclude that for a diatomic gas at constant volume, path 2 is related with a diatomic gas. From condition, $T_m>T_D,$we can say that is related with an isochoric process.

The molecule of diatomic gas is rotating because $T_m>T_D$, because some of the heat Q provided is going to the rotational energy in the diatomic gas molecule.

Thus, the molecules of diatomic gases are rotating.