One mole of and ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, `27^(@)C` . The work done on the gas will be :

Options

A. 300 R ln 7
B. 300 R ln 2
C. 300 R
D. 300 R ln 6

Answer

To solve the problem of calculating the work done on one mole of an ideal monoatomic gas compressed isothermally in a rigid vessel to double its pressure at room temperature (27°C), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Isothermal Process**:
In an isothermal process, the temperature remains constant. For an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by the ideal gas law:
\[ PV = nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin.

2. **Convert Temperature to Kelvin**:
The given temperature is 27°C. To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

3. **Initial and Final Pressures**:
Let the initial pressure be \( P_1 \) and the final pressure be \( P_2 \). According to the problem, the pressure is doubled:
\[ P_2 = 2P_1 \]

4. **Work Done in Isothermal Process**:
The work done on the gas during an isothermal process can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since the volume is constant in a rigid vessel, we can relate the volumes to pressures using the ideal gas law:
\[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \]
Substituting \( P_2 = 2P_1 \):
\[ \frac{V_2}{V_1} = \frac{P_1}{2P_1} = \frac{1}{2} \]

5. **Substituting into Work Formula**:
Now we can substitute this back into the work done formula:
\[ W = -nRT \ln\left(\frac{1}{2}\right) \]
Since \( n = 1 \) mole and \( R \) is the universal gas constant, we have:
\[ W = -RT \ln\left(\frac{1}{2}\right) \]

6. **Calculate Work Done**:
Now substituting the values:
\[ W = -R \cdot 300 \cdot \ln\left(\frac{1}{2}\right) \]
Using the property of logarithms, \( \ln\left(\frac{1}{2}\right) = -\ln(2) \):
\[ W = 300R \ln(2) \]

7. **Final Result**:
Therefore, the work done on the gas is:
\[ W = 300R \ln(2) \]

### Final Answer:
The work done on the gas will be \( 300R \ln(2) \).

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