4 moles of an ideal monoatomic gas is heated isobarically so that its absolute temperature increases 2 times. Then the entropy increment of the gas in this process is

Options

A. 285.J/k
B. 42.5J/k
C. 57.5J/k
D. 76.5J/k

Explanation

To find the entropy increment of the gas during the isobaric heating process, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 4 moles
- Initial temperature (T₀) = T₀ (arbitrary initial temperature)
- Final temperature (T) = 2T₀ (temperature increases to twice the initial)

2. **Understand the Process:**
- The process is isobaric, meaning the pressure remains constant while the temperature changes.

3. **Use the Formula for Entropy Change:**
The change in entropy (ΔS) for an ideal gas during an isobaric process can be calculated using the formula:
\[ \Delta S = n C_p \ln\left(\frac{T_f}{T_i}\right) \]
where \(C_p\) is the molar heat capacity at constant pressure.

4. **Determine \(C_p\) for a Monoatomic Ideal Gas:**
For a monoatomic ideal gas, the molar heat capacity at constant pressure is given by:
\[ C_p = \frac{5}{2} R \]
where \(R\) is the universal gas constant (approximately 8.314 J/(mol·K)).

5. **Calculate the Change in Entropy:**
Substitute the values into the entropy change formula:
\[ \Delta S = n C_p \ln\left(\frac{T_f}{T_i}\right) = 4 \times \left(\frac{5}{2} R\right) \ln\left(\frac{2T_0}{T_0}\right) \]
Simplifying gives:
\[ \Delta S = 4 \times \left(\frac{5}{2} R\right) \ln(2) \]
\[ \Delta S = 10 R \ln(2) \]

6. **Substitute the Value of \(R\):**
Now, substitute \(R\) into the equation:
\[ \Delta S = 10 \times 8.314 \times \ln(2) \]

7. **Calculate \(\ln(2)\):**
The value of \(\ln(2)\) is approximately 0.693.

8. **Final Calculation:**
\[ \Delta S = 10 \times 8.314 \times 0.693 \approx 57.5 \text{ J/K} \]

### Final Answer:
The entropy increment of the gas in this process is approximately **57.5 J/K**.

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