When two mole of an ideal gas `(C​_(p.m.) = 5/2R)` heated from 300K to 600K at constant volume. The change in entropy of gas will be :-

Options

A. `3/2` Rln2
B. 3 Rln2
C. 5 Rln2
D. `5/2` Rln2

Explanation

To find the change in entropy of the ideal gas when it is heated from 300 K to 600 K at constant volume, we can follow these steps:

### Step 1: Identify the Given Values
- Number of moles (n) = 2 moles
- Specific heat at constant pressure (C_p) = \( \frac{5}{2} R \)
- Initial temperature (T1) = 300 K
- Final temperature (T2) = 600 K

### Step 2: Calculate C_v
Using the relation between C_p and C_v:
\[ C_v = C_p - R \]
Substituting the value of C_p:
\[ C_v = \frac{5}{2} R - R = \frac{5}{2} R - \frac{2}{2} R = \frac{3}{2} R \]

### Step 3: Use the Entropy Change Formula
For an ideal gas at constant volume, the change in entropy (ΔS) can be calculated using the formula:
\[ \Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) \]

### Step 4: Substitute the Known Values
Substituting the values into the formula:
\[ \Delta S = 2 \times \frac{3}{2} R \ln\left(\frac{600}{300}\right) \]

### Step 5: Simplify the Expression
Calculating the logarithm:
\[ \frac{600}{300} = 2 \]
Thus,
\[ \ln(2) \text{ is a constant value.} \]
Now substituting this back into the equation:
\[ \Delta S = 2 \times \frac{3}{2} R \ln(2) \]
\[ \Delta S = 3R \ln(2) \]

### Final Result
The change in entropy of the gas is:
\[ \Delta S = 3R \ln(2) \]