An ideal monoatomic gas occupies volume `10^(-3)m^(3)` at temperature `3K` and pressure `10^(3)`Pa. The internal energy of the gas is taken to be zero at this point. It undergoes the following cycle: The temperature is raised to `300K` at constant volume, the gas is then expanded adiabatically till the temperature is `3K` , followed by isothermal compression to the original volume . Plot the process on a `PV` diagram. Calculate (i) The work done and the heat transferred in each process and the internal energy at the end of each process, (ii) The thermal efficiency of the cycle.

Explanation

To solve the problem step by step, we will analyze each part of the cycle for the ideal monoatomic gas and calculate the required quantities.

### Step 1: Determine the number of moles of the gas
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P = 10^3 \, \text{Pa} \)
- \( V = 10^{-3} \, \text{m}^3 \)
- \( R = 8.314 \, \text{J/(mol K)} \) (universal gas constant)
- \( T = 3 \, \text{K} \)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} = \frac{(10^3)(10^{-3})}{(8.314)(3)} \approx \frac{1}{24.942} \approx 0.0401 \, \text{mol} \]

### Step 2: Analyze the processes in the cycle
1. **Process AB (Constant Volume Heating from 3K to 300K)**
- Work done \( W_{AB} = 0 \) (since volume is constant).
- Change in internal energy \( \Delta U_{AB} = nC_v \Delta T \)
- For a monoatomic gas, \( C_v = \frac{3}{2} R \)
- \( \Delta T = 300 - 3 = 297 \, \text{K} \)
- \( \Delta U_{AB} = n \cdot \frac{3}{2} R \cdot (300 - 3) \)
- \( \Delta U_{AB} = 0.0401 \cdot \frac{3}{2} \cdot 8.314 \cdot 297 \approx 148.5 \, \text{J} \)
- Heat added \( Q_{AB} = \Delta U_{AB} = 148.5 \, \text{J} \)

2. **Process BC (Adiabatic Expansion from 300K to 3K)**
- Work done \( W_{BC} = -\Delta U_{BC} \) (since \( Q = 0 \))
- Using the adiabatic relation:
\[ W_{BC} = nC_v(T_1 - T_2) = n \cdot \frac{3}{2} R (300 - 3) \]
\[ W_{BC} = 0.0401 \cdot \frac{3}{2} \cdot 8.314 \cdot (300 - 3) \approx -148.5 \, \text{J} \]
- Since \( Q = 0 \), \( \Delta U_{BC} = -W_{BC} = 148.5 \, \text{J} \)

3. **Process CA (Isothermal Compression back to original volume)**
- Work done \( W_{CA} = nRT \ln \left(\frac{V_A}{V_C}\right) \)
- Since \( V_A = V_C = 10^{-3} \, \text{m}^3 \), the work done is:
\[ W_{CA} = nRT \ln \left(1\right) = 0 \, \text{J} \]
- Heat removed \( Q_{CA} = -W_{CA} = 0 \, \text{J} \)
- Change in internal energy \( \Delta U_{CA} = 0 \)

### Step 3: Calculate the thermal efficiency of the cycle
The thermal efficiency \( \eta \) is given by:
\[ \eta = \frac{Q_{\text{net}}}{Q_{\text{supply}}} \]
Where:
- \( Q_{\text{net}} = Q_{AB} - |Q_{CA}| = 148.5 - 0 = 148.5 \, \text{J} \)
- \( Q_{\text{supply}} = Q_{AB} = 148.5 \, \text{J} \)

Thus,
\[ \eta = \frac{148.5}{148.5} = 1 \text{ or } 100\% \]

### Summary of Results
1. **Work Done and Heat Transfer:**
- \( W_{AB} = 0 \, \text{J}, Q_{AB} = 148.5 \, \text{J}, \Delta U_{AB} = 148.5 \, \text{J} \)
- \( W_{BC} = -148.5 \, \text{J}, Q_{BC} = 0 \, \text{J}, \Delta U_{BC} = -148.5 \, \text{J} \)
- \( W_{CA} = 0 \, \text{J}, Q_{CA} = 0 \, \text{J}, \Delta U_{CA} = 0 \)

2. **Thermal Efficiency:**
- \( \eta = 100\% \)