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 In y`
B. `w = -300 x R In y/10y`
C. `w = - 300 x R ln 10`
D. `w = 100 x R In 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 Values:**
- Number of moles (n) = x moles
- Initial volume (V1) = y
- Final volume (V2) = 10y
- Temperature (T) = 27°C = 27 + 273 = 300 K

2. **Understand the Work Done Formula:**
- For an isothermal and reversible expansion of an ideal gas, the work done (W) is given by the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
- Here, R is the universal gas constant.

3. **Substitute the Known Values:**
- Substitute n = x, R (approximately 8.314 J/(mol·K)), T = 300 K, V2 = 10y, and V1 = y into the formula:
\[ W = -x \cdot R \cdot 300 \cdot \ln\left(\frac{10y}{y}\right) \]

4. **Simplify the Logarithm:**
- The logarithm simplifies as follows:
\[ \ln\left(\frac{10y}{y}\right) = \ln(10) \]
- Therefore, the equation for work done becomes:
\[ W = -x \cdot R \cdot 300 \cdot \ln(10) \]

5. **Final Expression:**
- Substitute R into the equation:
\[ W = -300xR \ln(10) \]
- This is the final expression for the work done during the isothermal expansion.

### Final Answer:
\[ W = -300xR \ln(10) \]