Question

A certain amount of energy is to be transferred as heat to 1 mol of a monatomic gas (a) at constant pressure and (b) at constant volume, and to 1 mol of a diatomic gas (c) at constant pressure and (d) at constant volume. Figure 19-19 shows four paths from an initial point to four final points on a p-v diagram for the two gases. Which path goes with which process? (e) Are the molecules of the diatomic gas rotating?

Short answer

1. For 1mol of monatomic gas at constant pressure, path 3 goes with isobaric process.
2. For 1mol of monatomic gas at constant volume, path 1 goes with isochoric process.
3. For 1mol of diatomic gas at constant pressure, path 4 goes with isobaric process.
4. For 1mol of diatomic gas at constant volume, path 2 goes with isochoric process.
5. Diatomic gas molecules are rotating.

Step-by-step solution

Stating the given data

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

Understanding the concept of internal energy and specific heat

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, $W=\underset{v_i}{\overset{v_f}{}}pdv$ …(i)

Change in internal energy at constant volume,$\Delta E_{int}=n{C}_v\Delta T$ …(ii)

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

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

$\Delta E=Q-W$ …(iv)

$\text{ΔE}={\text{nC}}_v\text{ΔT}$

(a) Calculation of the path of monatomic gas at constant pressure

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, $W=p\Delta V.$

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

$$\begin{gathered} n{C}_v\Delta T=Q-p\Delta V \\ Q=n{C}_v\Delta T+p\Delta V \end{gathered}$$ …(v)

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

$\Delta T=\frac{p\Delta V}{nR}$

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

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

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

$$\begin{gathered} Q=\left(\frac{\frac{3}{2}R}{R}+1\right)p\Delta V_m(\text{where subscript m is for monoatomic gas)} \\ =\frac{5}{2}p\Delta V_m \end{gathered}$$ …(vii)

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

$$\begin{gathered} Q=\left(\frac{\frac{5}{2}R}{R}+1\right)p\Delta V_d(\text{where subscript d is for diatomic gas)} \\ =\frac{7}{2}p\Delta V_d \end{gathered}$$ …(viii)

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

$$\begin{gathered} \frac{5}{2}p\Delta V_m=\frac{7}{2}p\Delta V_d \\ 5V_m=7V_d \end{gathered}$$

That means the volume of monatomic gas is greater than the volume of diatomic gas$\Delta V_m>\Delta V_d$

For any constant volume process

$W=0$

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

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

$Q=n\frac{3}{2}R\Delta T$ …(ix)

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

$Q=n\frac{5}{2}R\Delta T$ …(x)

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

$\begin{array}{c}n\frac{3}{2}R\Delta T_m=n\frac{5}{2}R\Delta T_d\\ 3\Delta T_m=5\Delta T_d\end{array}$

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 $\Delta V_m>\Delta V_d$, we can say that path 3 is related with an isobaric process.

(b) Calculation of the path of monatomic gas at constant volume

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.

(c) Calculation of the path of diatomic gas at constant pressure

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.

(d) Calculation of the path of diatomic gas at constant volume

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.

(e) Calculation of the behavior of the diatomic gas

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.