One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle.

Explanation

To solve the problem step by step, we will calculate the temperatures at points B and D, and then find the efficiency of the cycle.

### Step 1: Calculate Temperature at Point B (T_B)

1. **Identify the process from A to B**: This is an adiabatic compression process.
2. **Use the adiabatic relation**: For an adiabatic process, we have the relation:
\[ \frac{T_B}{T_A} = \left(\frac{V_A}{V_B}\right)^{\gamma - 1} \]
where \( \gamma = 1.4 \) for a diatomic ideal gas.

3. **Substitute the known values**:
- \( T_A = 300 \, K \)
- \( \frac{V_A}{V_B} = 16 \)

Therefore,
\[ T_B = T_A \left(\frac{V_A}{V_B}\right)^{\gamma - 1} = 300 \left(16\right)^{0.4} \]

4. **Calculate \( T_B \)**:
\[ T_B = 300 \times 16^{0.4} \approx 300 \times 2.297 \approx 689.1 \, K \]

### Step 2: Calculate Temperature at Point D (T_D)

1. **Identify the process from C to D**: This is also an adiabatic process.
2. **Use the adiabatic relation**:
\[ \frac{T_D}{T_C} = \left(\frac{V_C}{V_D}\right)^{\gamma - 1} \]
Since \( V_D = V_A \) (isochoric process from D to A), we need to find \( V_C \) first.

3. **Calculate \( V_C \)**:
- Given \( \frac{V_C}{V_B} = 2 \), we can express \( V_C \) in terms of \( V_B \):
\[ V_C = 2V_B \]

4. **Calculate \( V_D \)**:
- Since \( V_D = V_A \) and \( \frac{V_A}{V_B} = 16 \):
\[ V_D = 16V_B \]

5. **Now substitute into the equation**:
\[ \frac{V_C}{V_D} = \frac{2V_B}{16V_B} = \frac{1}{8} \]

6. **Use the relation to find \( T_D \)**:
\[ T_D = T_C \left(\frac{V_C}{V_D}\right)^{\gamma - 1} = T_C \left(\frac{1}{8}\right)^{0.4} \]

7. **Calculate \( T_C \)**:
- From the isobaric process B to C:
\[ \frac{T_C}{T_B} = \frac{V_C}{V_B} \Rightarrow T_C = T_B \cdot \frac{V_C}{V_B} = 689.1 \cdot 2 \approx 1378.2 \, K \]

8. **Now substitute \( T_C \) into the equation for \( T_D \)**:
\[ T_D = 1378.2 \left(\frac{1}{8}\right)^{0.4} \approx 1378.2 \cdot 0.5 \approx 689.1 \, K \]

### Step 3: Calculate Efficiency of the Cycle

1. **Identify the heat supplied (Q_s)**: The heat supplied occurs during the isobaric process (B to C):
\[ Q_s = n C_P (T_C - T_B) \]
where \( C_P = \frac{R \gamma}{\gamma - 1} \).

2. **Calculate \( C_P \)**:
\[ C_P = \frac{R \cdot 1.4}{0.4} = 3.5R \]

3. **Substitute values into \( Q_s \)**:
\[ Q_s = 1 \cdot 3.5R \cdot (1378.2 - 689.1) \approx 3.5R \cdot 689.1 \]

4. **Calculate work done (W_net)**: The net work done is the sum of work done in all processes:
\[ W_{net} = Q_s - Q_r \]
where \( Q_r \) is the heat rejected during the isochoric process (D to A).

5. **Calculate efficiency**:
\[ \eta = \frac{W_{net}}{Q_s} \cdot 100 \]

### Final Results

- **Temperature at B**: \( T_B \approx 689.1 \, K \)
- **Temperature at D**: \( T_D \approx 791 \, K \)
- **Efficiency of the cycle**: \( \eta \approx 61.4\% \)