P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be

Options

A. 4R
B. `2.5R`
C. 3R
D. `(4R)/(3)`

Answer

To solve the problem regarding the molar heat capacity of a diatomic gas in a process represented by a straight line on a P-V diagram, we can follow these 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 indicates a linear relationship between pressure (P) and volume (V).

**Hint:** A straight line through the origin implies that P is directly proportional to V.

### Step 2: Apply the First Law of Thermodynamics
According to the first law of thermodynamics, we have:
\[ Q = U + W \]
Where:
- \( Q \) is the heat added to the system,
- \( U \) is the internal energy,
- \( W \) is the work done by the system.

**Hint:** Remember that the internal energy change \( U \) for an ideal gas can be expressed in terms of molar heat capacity.

### Step 3: Express the Work Done
For a process represented by a straight line on a P-V diagram, the work done \( W \) can be calculated as the area under the curve. Since it forms a triangle, the area can be calculated as:
\[ W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times V \times P \]

**Hint:** The area of a triangle is half the product of its base and height.

### Step 4: Relate Pressure, Volume, and Temperature
Using the ideal gas law:
\[ PV = nR\Delta T \]
This can be substituted into the work done expression.

**Hint:** Use the ideal gas law to express \( PV \) in terms of temperature.

### Step 5: Substitute into the First Law Equation
Substituting the expressions for \( Q \), \( U \), and \( W \) into the first law equation gives:
\[ nC\Delta T = nC_v\Delta T + \frac{nR\Delta T}{2} \]

**Hint:** Factor out common terms to simplify the equation.

### Step 6: Simplify the Equation
Taking \( n\Delta T \) common from both sides, we get:
\[ C = C_v + \frac{R}{2} \]

**Hint:** Isolate \( C \) to find the relationship between heat capacities.

### Step 7: Substitute the Value of \( C_v \)
For a diatomic gas, the molar heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{5}{2}R \]
Substituting this into the equation gives:
\[ C = \frac{5}{2}R + \frac{R}{2} = 3R \]

**Hint:** Ensure you perform the arithmetic correctly when combining terms.

### Step 8: Conclusion
Thus, the molar heat capacity \( C \) for the diatomic gas in this process is:
\[ C = 3R \]

**Final Answer:** The correct option is \( 3R \).

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