When one mole of an ideal gas is heated to twice of its initial temperature and simultaneously its pressure is increased to twice of its initial pressure, the change in entropy `(DeltaS)` is:

Options

A. `C_(p)ln2`
B. `R ln2`
C. `(C_(P)-R)ln2`
D. `(R-C_(p))ln2`

Answer

To solve the problem of finding the change in entropy (ΔS) when one mole of an ideal gas is heated to twice its initial temperature and its pressure is increased to twice its initial pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have 1 mole of an ideal gas.
- Initial temperature \( T_1 = T \).
- Final temperature \( T_2 = 2T \).
- Initial pressure \( P_1 = P \).
- Final pressure \( P_2 = 2P \).

2. **Use the Entropy Change Formula**:
The change in entropy for an ideal gas can be expressed as:
\[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right) \]
Since \( n = 1 \) (one mole), the equation simplifies to:
\[ \Delta S = C_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \]

3. **Substitute the Known Values**:
Substitute \( T_2 = 2T \), \( T_1 = T \), \( P_2 = 2P \), and \( P_1 = P \):
\[ \Delta S = C_p \ln\left(\frac{2T}{T}\right) - R \ln\left(\frac{2P}{P}\right) \]

4. **Simplify the Logarithmic Expressions**:
This simplifies to:
\[ \Delta S = C_p \ln(2) - R \ln(2) \]
Factor out \( \ln(2) \):
\[ \Delta S = \ln(2) \left(C_p - R\right) \]

5. **Use the Ideal Gas Relation**:
For an ideal gas, we know that:
\[ C_p - R = C_v \]
Therefore, we can express the change in entropy as:
\[ \Delta S = C_v \ln(2) \]

### Final Answer:
The change in entropy \( \Delta S \) is:
\[ \Delta S = C_v \ln(2) \]