91Ó°ÊÓ

1. Number of moles in the sample,$n=1.0\mathrm{mol}$
2. Volume of the gas,$V_1=4.00V_0$
3. P-V diagram for a monatomic ideal gas taken through the cycle.

We can find thevalue of$\mathit{W}\mathbf{/}{\mathit{p}}_{\mathbf{0}}{\mathit{V}}_{\mathbf{0}}$ as the gas goes from the state a to state c along path abc using the formula for work done by an ideal gas in terms of pressure and volume change. Then using the gas law, we can find the temperature of the gas at points b and c. Then using it in the formula for heat absorbed, we can find the amount of heat absorbed. Inserting it into the equation for the law of conservation of energy, we can find the value of $\mathbf{∆}{\mathit{E}}_{\mathbf{i}\mathbf{n}\mathbf{t}}\mathbf{}\mathbf{/}{\mathit{p}}_{\mathbf{0}}{\mathit{V}}_{\mathbf{0}}$. By using the formula for entropy change, we can find its value.

Formulae:

The work done by a gas at constant pressure, $W=p∆V$ …(¾±)

The ideal-gas equation, $PV=nRT$ …(¾±¾±)

The molar specific heat at constant volume, $C_V=\frac{3}{2}R$ …(¾±¾±¾±)

The internal energy of the gas from first law of thermodynamics, $∆E=Q-W$ …(¾±±¹)

The entropy change of the system, $∆S=n{C}_{V}In\frac{T_f}{T_i}$ …(±¹)

The heat absorbed by the body, $Q=n{C}_V∆T$ …(±¹¾±)

From the given p-V diagram,

Along path, the work done by the gas using equation (i) is given as:

$$\begin{gathered} W=p_0\left(V_1-V_0\right) \\ =4p_0\left(V_0-V_0\right) \\ =3p_0{V}_0 \end{gathered}$$

The required value is given as:

$\frac{W}{p_0{V}_0}=3$

Therefore, the value of $W/p_0{V}_0$as the gas goes from state a to state c along path abc is 3.

The temperature of gas at point b using equation (ii) is given as:

$$\begin{gathered} T_b=\frac{p_0{V}_1}{nR} \\ =\frac{4p_0{V}_0}{\left(1\right)R} \end{gathered}$$

The temperature of gas at point c using equation (ii) is given as:

$$\begin{gathered} T_c=\frac{2p_0{V}_1}{nR} \\ =\frac{8p_0{V}_0}{\left(1\right)R} \end{gathered}$$

The heat energy absorbed is given by substituting the value of equation (iii) in equation (vi) as follows:

$$\begin{gathered} Q=1×\frac{3}{2}R\left(T_c-T_b\right) \\ =1×\frac{3}{2}R\left(\frac{8p_0{V}_0}{\left(1\right)R}-\frac{4p_0{V}_0}{\left(1\right)R}\right) \\ =6p_0{V}_0 \end{gathered}$$

Since along path bc, the gas undergoes an isochoric process. So,$W=0$

The change in internal energy using equation (iv) can be given as:

$$\begin{gathered} ∆E=Q \\ =6p_0{V}_0 \\ \frac{∆E}{p_0{V}_0}=6 \end{gathered}$$

Therefore, the value of $∆E/p_0{V}_0$ as the gas goes from state b to state c along pathabc is 6.

Since energy is a state function,$∆E_{int}=0$

Therefore, the value of $∆E/p_0{V}_0$as the gas is going through one full cycle is zero.

The change in entropy along path bc using equation (v) and the given values is given as:

$$\begin{gathered} ∆S=1×\frac{3}{2}×8.314ln=\frac{{\displaystyle \frac{8p_0{V}_0}{\left(1\right)R}}}{{\displaystyle \frac{4p_0{V}_0}{\left(1\right)R}}} \\ =8.64\mathrm{J}/\mathrm{K} \end{gathered}$$

Therefore, the value of $∆S$in going from b to c is $8.64\mathrm{J}/\mathrm{K}$.

For a complete cycle,$T_i=T_f$

Substituting it in the above formula of equation (v) for entropy change gives$∆S=0$

Therefore, the value of $∆S$ through one full cycle is zero.