If x mole of ideal gas at `27^(@)C` expands isothermally and reversibly from a volume of y to 10 y, then the work done is

Options

A. `w = x R 300 ln y`
B. `w = -300 x R ln y/10y`
C. `w = - 300 x R ln 10`
D. `w = 100 x R ln 1/y`

Answer

To solve the problem of calculating the work done during the isothermal and reversible expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Number of moles of gas, \( n = x \)
- Initial volume, \( V_1 = y \)
- Final volume, \( V_2 = 10y \)
- Temperature, \( T = 27^\circ C \)

2. **Convert Temperature to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T = 27 + 273 = 300 \, K \]

3. **Use the Formula for Work Done in Isothermal Expansion:**
- The work done during an isothermal reversible expansion is given by the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
- Here, \( R \) is the universal gas constant.

4. **Substitute the Values into the Formula:**
- Substitute \( n = x \), \( R \) (approximately \( 8.314 \, J/(mol \cdot K) \)), \( T = 300 \, K \), \( V_2 = 10y \), and \( V_1 = y \):
\[ W = -x \cdot R \cdot 300 \cdot \ln\left(\frac{10y}{y}\right) \]

5. **Simplify the Logarithmic Term:**
- Since \( \frac{10y}{y} = 10 \):
\[ W = -x \cdot R \cdot 300 \cdot \ln(10) \]

6. **Final Expression for Work Done:**
- Thus, the work done \( W \) can be expressed as:
\[ W = -300xR \ln(10) \]

### Final Answer:
The work done during the isothermal and reversible expansion of the ideal gas is:
\[ W = -300xR \ln(10) \]