When two moles of an ideal gas `(C_(p.m.)=(5)/(2)R)` heated form 300K to 600K at constant pressure, the change in entropy of gas `(DeltaS)` is:

Options

A. `(3)/(2) R ln 2`
B. `-(3)/(2) R ln 2`
C. `5R ln 2`
D. `(5)/(2) R ln 2`

Explanation

To solve the problem of finding the change in entropy (ΔS) of an ideal gas when heated at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Number of moles (N) = 2 moles
- Heat capacity at constant pressure (C_p) = \( \frac{5}{2} R \)
- Initial temperature (T1) = 300 K
- Final temperature (T2) = 600 K

2. **Use the Formula for Change in Entropy**:
The formula for the change in entropy (ΔS) for an ideal gas at constant pressure is given by:
\[ \Delta S = N C_p \ln\left(\frac{T_2}{T_1}\right) \]
Since the pressure is constant, we can ignore the pressure terms in the formula.

3. **Substitute the Known Values into the Formula**:
\[ \Delta S = 2 \times \left(\frac{5}{2} R\right) \ln\left(\frac{600}{300}\right) \]

4. **Simplify the Expression**:
- Calculate \( \frac{600}{300} = 2 \).
- Therefore, the expression becomes:
\[ \Delta S = 2 \times \left(\frac{5}{2} R\right) \ln(2) \]

5. **Further Simplification**:
- The 2s cancel out:
\[ \Delta S = 5 R \ln(2) \]

6. **Final Result**:
The change in entropy (ΔS) of the gas is:
\[ \Delta S = 5 R \ln(2) \]