When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas `(DelataS)` is:

Options

A. `C_(p,m)ln2`
B. `C_(v,m)ln2`
C. `R ln 2`
D. `(C_(v,m)-R)ln2`

Explanation

To find the change in entropy (ΔS) of one mole of an ideal gas that is compressed to half of its initial volume and heated to twice its initial temperature, we can use the formula for the change in entropy:

\[ \Delta S = nC_V \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right) \]

### Step-by-Step Solution:

1. **Identify the variables:**
- Given that \( n = 1 \) mole (since we are dealing with one mole of gas).
- Let the initial temperature be \( T_1 = T \).
- The final temperature is \( T_2 = 2T \) (since it is heated to twice its initial temperature).
- Let the initial volume be \( V_1 = V \).
- The final volume is \( V_2 = \frac{V}{2} \) (since it is compressed to half its initial volume).

2. **Substitute the values into the entropy formula:**
\[ \Delta S = 1 \cdot C_V \ln\left(\frac{2T}{T}\right) + 1 \cdot R \ln\left(\frac{\frac{V}{2}}{V}\right) \]

3. **Simplify the logarithmic terms:**
- The first term simplifies to:
\[ \Delta S_1 = C_V \ln(2) \]
- The second term simplifies to:
\[ \Delta S_2 = R \ln\left(\frac{1}{2}\right) = -R \ln(2) \]

4. **Combine the two terms:**
\[ \Delta S = C_V \ln(2) - R \ln(2) \]
\[ \Delta S = (C_V - R) \ln(2) \]

5. **Final expression for the change in entropy:**
\[ \Delta S = (C_V - R) \ln(2) \]

### Conclusion:
The change in entropy of the gas is given by:
\[ \Delta S = (C_V - R) \ln(2) \]