Three moles of an ideal gas being initially at a temperature `T_i=273K` were isothermally expanded 5 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total heat supplied in the process is 80kJ. Find `gamma(=(C_p)/(C_V))` of the gas.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the processes involved
We have two processes:
1. Isothermal expansion from state A to state B.
2. Isochoric heating from state B to state C.

### Step 2: Analyze the isothermal expansion (A to B)
- Initial temperature \( T_i = 273 \, K \)
- Initial volume \( V_A = V_0 \)
- Final volume after expansion \( V_B = 5V_0 \)
- Since the process is isothermal, the temperature remains constant, and we can use the ideal gas law.

Using the ideal gas law:
\[ P_A V_A = nRT_A \quad \text{(Initial state)} \]
\[ P_B V_B = nRT_B \quad \text{(Final state)} \]
Since \( T_A = T_B \) (isothermal), we can write:
\[ \frac{P_B}{P_A} = \frac{V_A}{V_B} = \frac{V_0}{5V_0} = \frac{1}{5} \]
Thus,
\[ P_B = \frac{P_A}{5} \]

### Step 3: Analyze the isochoric heating (B to C)
- The volume remains constant during this process, so \( V_C = V_B = 5V_0 \).
- The pressure at state C must equal the initial pressure \( P_A \).

Using the ideal gas law again:
\[ P_B V_B = nRT_B \quad \text{(State B)} \]
\[ P_C V_C = nRT_C \quad \text{(State C)} \]
Since \( P_C = P_A \) and \( V_C = V_B \):
\[ P_A \cdot 5V_0 = nRT_C \]
From the previous step, we know \( P_B = \frac{P_A}{5} \), so:
\[ \frac{P_A}{5} \cdot 5V_0 = nRT_B \implies P_A V_0 = nRT_B \]
Thus, \( T_B = T_A = 273 \, K \).

### Step 4: Relate temperatures
From the isochoric heating:
\[ \frac{P_C}{P_B} = \frac{T_C}{T_B} \implies \frac{P_A}{\frac{P_A}{5}} = \frac{T_C}{T_B} \implies 5 = \frac{T_C}{T_B} \implies T_C = 5T_B = 5 \times 273 = 1365 \, K \]

### Step 5: Calculate heat transfer
1. **Heat transfer during isothermal expansion (A to B)**:
\[ Q_{AB} = W_{AB} = nRT \ln \left( \frac{V_B}{V_A} \right) = nRT_A \ln(5) \]
Where \( n = 3 \) moles and \( R = 8.31 \, J/(mol \cdot K) \):
\[ Q_{AB} = 3 \times 8.31 \times 273 \ln(5) \]

2. **Heat transfer during isochoric heating (B to C)**:
\[ Q_{BC} = \Delta U = nC_V \Delta T = nC_V (T_C - T_B) \]
Where \( \Delta T = T_C - T_B = 1365 - 273 = 1092 \, K \).

### Step 6: Total heat supplied
The total heat supplied is given as:
\[ Q_{total} = Q_{AB} + Q_{BC} = 80 \times 10^3 \, J \]

### Step 7: Substitute and solve for \( \gamma \)
Using the relation for \( C_V \):
\[ C_V = \frac{R}{\gamma - 1} \]
Substituting \( Q_{AB} \) and \( Q_{BC} \) into the total heat equation:
\[ 3 \times 8.31 \times 273 \ln(5) + 3 \times \frac{R}{\gamma - 1} \times 1092 = 80 \times 10^3 \]
Now, we can solve for \( \gamma \).

### Step 8: Solve for \( \gamma \)
1. Calculate \( Q_{AB} \):
\[ Q_{AB} = 3 \times 8.31 \times 273 \ln(5) \approx 3 \times 8.31 \times 273 \times 1.609 = 3 \times 8.31 \times 439.8 \approx 10958.7 \, J \]

2. Substitute into the total heat equation:
\[ 10958.7 + \frac{3 \times 1092 \times 8.31}{\gamma - 1} = 80000 \]
Rearranging gives:
\[ \frac{3 \times 1092 \times 8.31}{\gamma - 1} = 80000 - 10958.7 \]

3. Solve for \( \gamma \):
\[ \frac{3 \times 1092 \times 8.31}{\gamma - 1} = 68941.3 \]
\[ \gamma - 1 = \frac{3 \times 1092 \times 8.31}{68941.3} \]
\[ \gamma = 1 + \frac{3 \times 1092 \times 8.31}{68941.3} \]

Finally, calculating the value gives \( \gamma \approx 1.4 \).

### Final Answer
\[ \gamma \approx 1.4 \]