Four moles of a certain ideal gas at `30^@C` are expanded isothermally to three times its volume and then heated at this constant volume until the pressure is raised to its initial value. In the whole process the heat supplied is 72 KJ. Calculate the ratio `C_p//C_v` for the gas and state whether it is monoatomic, diatomic or polyatomic gas.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Process
We have 4 moles of an ideal gas at an initial temperature of \(30^\circ C\) (which is \(303 K\)). The gas is expanded isothermally to three times its initial volume and then heated at constant volume until the pressure returns to its initial value. The total heat supplied during this process is \(72 \, \text{kJ}\).

### Step 2: Calculate the Work Done During Isothermal Expansion
For an isothermal process, the work done \(W\) by the gas when it expands from volume \(V_i\) to \(V_f\) is given by:
\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]
Here, \(V_f = 3V_i\), so:
\[ W = nRT \ln(3) \]
Substituting \(n = 4\) moles, \(R = 8.31 \, \text{J/(mol K)}\), and \(T = 303 \, K\):
\[ W = 4 \times 8.31 \times 303 \ln(3) \]
Calculating \(W\):
\[ W \approx 4 \times 8.31 \times 303 \times 1.0986 \approx 1016.7 \, \text{J} \approx 1.0167 \, \text{kJ} \]

### Step 3: Calculate the Heat Supplied in the Isochoric Process
In the second part of the process (constant volume), the heat supplied \(Q_2\) can be calculated using the change in internal energy:
\[ Q_2 = \Delta U = nC_v \Delta T \]
Since the pressure returns to its initial value, we can use the ideal gas law to find the final temperature \(T_c\) after heating:
Using the relation \(P_1/T_1 = P_2/T_2\) at constant volume:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_c} \implies T_c = T_1 \cdot \frac{P_2}{P_1} \]
Since \(P_2 = P_1\), we have:
\[ T_c = 3T_1 = 3 \times 303 = 909 \, K \]
Now, we can find \(\Delta T\):
\[ \Delta T = T_c - T_1 = 909 - 303 = 606 \, K \]
Thus, the heat supplied during the constant volume process is:
\[ Q_2 = nC_v \Delta T = 4C_v \times 606 \]

### Step 4: Total Heat Supplied
The total heat supplied is:
\[ Q = Q_1 + Q_2 \]
Where \(Q_1 = W\) and \(Q_2 = 4C_v \times 606\):
\[ 72 \, \text{kJ} = 1.0167 \, \text{kJ} + 4C_v \times 606 \]
Converting \(72 \, \text{kJ}\) to J:
\[ 72000 \, \text{J} = 1016.7 \, \text{J} + 4C_v \times 606 \]
Solving for \(C_v\):
\[ 72000 - 1016.7 = 4C_v \times 606 \]
\[ 70983.3 = 4C_v \times 606 \]
\[ C_v = \frac{70983.3}{4 \times 606} \approx 29.2 \, \text{J/(mol K)} \]

### Step 5: Calculate \(C_p\) and the Ratio \(C_p/C_v\)
Using the relation \(C_p = C_v + R\):
\[ C_p = 29.2 + 8.31 \approx 37.51 \, \text{J/(mol K)} \]
Now, we can find the ratio:
\[ \frac{C_p}{C_v} = \frac{37.51}{29.2} \approx 1.28 \]

### Step 6: Determine the Type of Gas
The degrees of freedom \(f\) can be calculated using the relation:
\[ C_v = \frac{f}{2}R \]
From \(C_v = 29.2\):
\[ 29.2 = \frac{f}{2} \times 8.31 \implies f \approx 7 \]
Since \(f = 6\) corresponds to a polyatomic gas, we conclude that the gas is polyatomic.

### Final Answer
The ratio \(\frac{C_p}{C_v} \approx 1.28\) and the gas is polyatomic.

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