If the P-V diagram of a diatomic gas is plotted, it is a straight line passing through the origin. The molar heat capacity of the gas in the process is IR, where I is an integer. Find the value of I.

Explanation

To solve the problem, we need to analyze the given information about the diatomic gas and its behavior in a polytropic process. Let's break down the steps:

### Step 1: Understand the P-V Diagram
The problem states that the P-V diagram of the diatomic gas is a straight line passing through the origin. This implies that pressure (P) is directly proportional to volume (V). Mathematically, we can express this as:
\[ P = kV \]
for some constant \( k \).

### Step 2: Identify the Type of Process
Since the P-V relationship is linear, we can relate it to a polytropic process where:
\[ PV^n = \text{constant} \]
Here, the exponent \( n \) can be determined from the linear relationship. For our case, since \( P \) is proportional to \( V \), we can conclude that:
\[ n = -1 \]

### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
For a polytropic process, we can express the heat added \( Q \) in terms of the heat capacity \( C \):
\[ Q = nC\Delta T \]
where \( C \) is the molar heat capacity.

### Step 4: Relate Internal Energy Change and Work Done
The change in internal energy \( \Delta U \) for an ideal gas can be expressed as:
\[ \Delta U = nC_v\Delta T \]
where \( C_v \) is the molar heat capacity at constant volume.

The work done \( W \) in a polytropic process can be expressed as:
\[ W = \frac{nR\Delta T}{1 - n} \]
Substituting \( n = -1 \), we get:
\[ W = \frac{nR\Delta T}{1 - (-1)} = \frac{nR\Delta T}{2} \]

### Step 5: Substitute Values into the First Law
Now substituting these expressions into the first law equation:
\[ nC\Delta T = nC_v\Delta T + \frac{nR\Delta T}{2} \]
Dividing through by \( n\Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ C = C_v + \frac{R}{2} \]

### Step 6: Calculate Molar Heat Capacity for Diatomic Gas
For a diatomic gas, the molar heat capacity at constant volume \( C_v \) is given by:
\[ C_v = \frac{5R}{2} \]
Now substituting this into our equation for \( C \):
\[ C = \frac{5R}{2} + \frac{R}{2} = \frac{6R}{2} = 3R \]

### Step 7: Identify the Value of I
The problem states that the molar heat capacity is \( IR \). From our calculation, we found:
\[ C = 3R \]
Thus, we can equate:
\[ IR = 3R \]
This gives us:
\[ I = 3 \]

### Final Answer
The value of \( I \) is:
\[ \boxed{3} \]