One mole of an ideal gas is originally at p₀, V₀, and T₀ . The gas is heated at constant volume to 2T₀, then allowed to expand at constant temperature to 2V₀, and finally it is allowed a cool at constant pressure to T₀, the net entropy change for this ideal gas is (A)∆S = (5R/2) In 2.

(B)∆S=5R/2.

(C)∆S = R In 2.

(D)∆S=3R/2.

(E) ∆S = 0.

To determine the net entropy change for the ideal gas undergoing the specified processes, we can break down each step and analyze the changes in entropy associated with them. The processes described are heating at constant volume, isothermal expansion, and cooling at constant pressure. Let's go through each step systematically.

Step 1: Heating at Constant Volume

When the gas is heated at constant volume from an initial temperature \( T_0 \) to a final temperature \( 2T_0 \), the change in entropy (\( \Delta S_1 \)) can be calculated using the formula:

• Formula: \( \Delta S = nC_v \ln\left(\frac{T_f}{T_i}\right) \)

For one mole of an ideal gas, the molar heat capacity at constant volume \( C_v \) is \( \frac{3R}{2} \) for a monatomic gas. Thus, we have:

• Initial temperature \( T_i = T_0 \)
• Final temperature \( T_f = 2T_0 \)

Substituting these values into the formula gives:

• \( \Delta S_1 = 1 \cdot \frac{3R}{2} \ln\left(\frac{2T_0}{T_0}\right) = \frac{3R}{2} \ln(2) \)

Step 2: Isothermal Expansion

Next, the gas expands isothermally from volume \( V_0 \) to \( 2V_0 \) at a constant temperature of \( 2T_0 \). The change in entropy for this process (\( \Delta S_2 \)) can be calculated using:

• Formula: \( \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \)

Here, \( V_i = V_0 \) and \( V_f = 2V_0 \). Thus, we have:

• \( \Delta S_2 = 1 \cdot R \ln\left(\frac{2V_0}{V_0}\right) = R \ln(2) \)

Step 3: Cooling at Constant Pressure

Finally, the gas cools at constant pressure back to the original temperature \( T_0 \). The change in entropy for this process (\( \Delta S_3 \)) is given by:

• Formula: \( \Delta S = nC_p \ln\left(\frac{T_f}{T_i}\right) \)

For one mole of an ideal gas, the molar heat capacity at constant pressure \( C_p \) is \( \frac{5R}{2} \). Here, \( T_i = 2T_0 \) and \( T_f = T_0 \), so:

• \( \Delta S_3 = 1 \cdot \frac{5R}{2} \ln\left(\frac{T_0}{2T_0}\right) = \frac{5R}{2} \ln\left(\frac{1}{2}\right) = -\frac{5R}{2} \ln(2) \)

Calculating the Total Entropy Change

Now, we can find the total change in entropy (\( \Delta S_{total} \)) by summing the changes from each step:

• \( \Delta S_{total} = \Delta S_1 + \Delta S_2 + \Delta S_3 \)
• \( \Delta S_{total} = \frac{3R}{2} \ln(2) + R \ln(2) - \frac{5R}{2} \ln(2) \)

Combining these terms gives:

• \( \Delta S_{total} = \left(\frac{3R}{2} + R - \frac{5R}{2}\right) \ln(2) = 0 \)

Final Result

The net entropy change for the entire process is:

• Answer: \( \Delta S_{total} = 0 \)

Thus, the correct option is (E) \( \Delta S = 0 \). This indicates that the overall process is reversible, as the entropy change for the system returns to its original state after completing the cycle.