The entropy change when two moles of ideal monatomic gas is heated from `200 "to" 300^(@) C` reversibly and isochorically ?

Options

A. `(3)/(2) R In ((300)/(200))`
B. `(5)/(2) R In ((573)/(273))`
C. `3R In ((573)/(473))`
D. `(3)/(2) R In ((573)/(473))`

Explanation

To find the entropy change when two moles of an ideal monatomic gas is heated from 200°C to 300°C reversibly and isochorically, we can follow these steps:

### Step 1: Understand the Process
Since the process is isochoric (constant volume), the work done (W) is zero. According to the first law of thermodynamics, the heat added (Q) is equal to the change in internal energy (ΔU).

### Step 2: Write the Formula for Entropy Change
The change in entropy (ΔS) for a reversible process can be expressed as:
\[ \Delta S = \int \frac{dQ}{T} \]
For an isochoric process, we can express the heat added as:
\[ dQ = n C_V dT \]
where \(C_V\) is the molar heat capacity at constant volume.

### Step 3: Set Up the Integral
Substituting \(dQ\) into the entropy change formula, we have:
\[ \Delta S = \int_{T_1}^{T_2} \frac{n C_V dT}{T} \]

### Step 4: Identify Values
Given:
- \(n = 2\) moles
- For a monatomic ideal gas, \(C_V = \frac{3}{2}R\)
- Convert temperatures from Celsius to Kelvin:
- \(T_1 = 200°C + 273 = 473 K\)
- \(T_2 = 300°C + 273 = 573 K\)

### Step 5: Perform the Integration
Substituting the values into the integral:
\[ \Delta S = n C_V \int_{T_1}^{T_2} \frac{dT}{T} = n C_V \left[ \ln T \right]_{T_1}^{T_2} \]
\[ \Delta S = n C_V (\ln T_2 - \ln T_1) = n C_V \ln \frac{T_2}{T_1} \]

### Step 6: Substitute Known Values
Substituting the known values:
\[ \Delta S = 2 \cdot \frac{3}{2}R \ln \frac{573}{473} \]
\[ \Delta S = 3R \ln \frac{573}{473} \]

### Final Result
Thus, the change in entropy when two moles of an ideal monatomic gas is heated from 200°C to 300°C reversibly and isochorically is:
\[ \Delta S = 3R \ln \frac{573}{473} \]