Two moles of helium gas`(lambda=5//3)` are initially at temperature `27^@C` and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(i) Sketch the process on a p-V diagram.
(ii) What are the final volume and pressure of the gas?
(iii) What is the work done by the gas ?

Explanation

Let's solve the problem step by step.

### Given Data:
- Number of moles of helium gas, \( n = 2 \)
- Initial temperature, \( T_1 = 27^\circ C = 300 \, K \)
- Initial volume, \( V_1 = 20 \, L = 20 \times 10^{-3} \, m^3 \)
- \( \lambda = \frac{5}{3} \)

### Step 1: Calculate Initial Pressure
Using the ideal gas equation:
\[ PV = nRT \]
We can rearrange it to find pressure:
\[ P_1 = \frac{nRT_1}{V_1} \]
Substituting the values:
\[ P_1 = \frac{2 \times 8.31 \, \text{J/(mol K)} \times 300 \, K}{20 \times 10^{-3} \, m^3} \]
Calculating:
\[ P_1 = \frac{4986}{0.02} = 249300 \, \text{Pa} = 2.493 \times 10^5 \, \text{N/m}^2 \]

### Step 2: Determine Final Volume After Constant Pressure Expansion
The gas expands at constant pressure until the volume is doubled:
\[ V_2 = 2V_1 = 2 \times 20 \times 10^{-3} = 40 \times 10^{-3} \, m^3 \]

### Step 3: Calculate Final Temperature After Expansion
Since the process is at constant pressure, we can use the relation \( V \propto T \):
\[ \frac{V_2}{V_1} = \frac{T_2}{T_1} \implies T_2 = 2T_1 = 2 \times 300 = 600 \, K \]

### Step 4: Adiabatic Process to Return to Initial Temperature
For the adiabatic process from state 2 to state 3, we use:
\[ T_2 V_2^{\lambda - 1} = T_3 V_3^{\lambda - 1} \]
Since \( T_3 = T_1 = 300 \, K \):
\[ 600 \times (40 \times 10^{-3})^{\frac{2}{3}} = 300 \times V_3^{\frac{2}{3}} \]
Rearranging gives:
\[ V_3^{\frac{2}{3}} = \frac{600 \times (40 \times 10^{-3})^{\frac{2}{3}}}{300} \]
Calculating \( V_3 \):
\[ V_3^{\frac{2}{3}} = 2 \times (40 \times 10^{-3})^{\frac{2}{3}} = 2 \times (40^{\frac{2}{3}}) \times (10^{-3})^{\frac{2}{3}} \]
Calculating \( 40^{\frac{2}{3}} \):
\[ 40^{\frac{2}{3}} \approx 16.00 \implies V_3 \approx (2 \times 16.00)^{\frac{3}{2}} \times 10^{-2} \approx 3.1 \times 10^{-3} \, m^3 \]

### Step 5: Calculate Final Pressure
Using the ideal gas equation again:
\[ P_3 = \frac{nRT_3}{V_3} \]
Substituting the values:
\[ P_3 = \frac{2 \times 8.31 \times 300}{3.1 \times 10^{-3}} \]
Calculating:
\[ P_3 \approx 0.44 \times 10^5 \, \text{N/m}^2 \]

### Step 6: Calculate Work Done by the Gas
The total work done is the sum of the work done in both processes:
1. Work done during constant pressure expansion (1 to 2):
\[ W_{1 \to 2} = P_1 (V_2 - V_1) = P_1 \times (40 \times 10^{-3} - 20 \times 10^{-3}) = P_1 \times 20 \times 10^{-3} \]
Substituting \( P_1 \):
\[ W_{1 \to 2} = 2.493 \times 10^5 \times 20 \times 10^{-3} = 4986 \, J \]

2. Work done during adiabatic process (2 to 3):
\[ W_{2 \to 3} = \frac{nR(T_3 - T_2)}{\lambda - 1} \]
Substituting:
\[ W_{2 \to 3} = \frac{2 \times 8.31 \times (300 - 600)}{\frac{5}{3} - 1} = \frac{2 \times 8.31 \times (-300)}{\frac{2}{3}} = -12479 \, J \]

### Total Work Done
\[ W_{total} = W_{1 \to 2} + W_{2 \to 3} = 4986 - 12479 = -7493 \, J \]

### Summary of Results
- Final Volume \( V_3 \approx 3.1 \times 10^{-3} \, m^3 \)
- Final Pressure \( P_3 \approx 0.44 \times 10^5 \, \text{N/m}^2 \)
- Work Done \( W_{total} \approx -7493 \, J \)