1 mole of an ideal gas expands from `5 dm^(3) " to" 25 dm^(3)` isothermally and irreversibly at `27^(@)` C. Find work done in the process [R=8.3 `J//mol//K`]

Options

A. `-1.99kJ`
B. `+1.99kJ`
C. `-7.46kJ`
D. `+7.46kJ`

Answer

To find the work done by 1 mole of an ideal gas that expands isothermally and irreversibly from 5 dm³ to 25 dm³ at a temperature of 27°C, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 27 + 273.15 = 300.15 \approx 300 \, K \]

### Step 2: Identify the initial and final volumes
The initial volume \( V_1 \) is 5 dm³ and the final volume \( V_2 \) is 25 dm³.

### Step 3: Use the ideal gas equation to find the final pressure
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( n = 1 \, \text{mol} \)
- \( R = 8.3 \, \text{J/mol·K} \)
- \( T = 300 \, K \)
- \( V_2 = 25 \, \text{dm}^3 = 25 \times 10^{-3} \, \text{m}^3 \) (since 1 dm³ = 0.001 m³)

We can rearrange the ideal gas equation to find the pressure \( P \):
\[ P = \frac{nRT}{V} \]
Substituting the values for \( V_2 \):
\[ P_2 = \frac{1 \times 8.3 \times 300}{25 \times 10^{-3}} \]
\[ P_2 = \frac{2490}{0.025} = 99600 \, \text{Pa} \, \text{or} \, 99.6 \, \text{kPa} \]

### Step 4: Calculate the work done using the formula for irreversible expansion
For an isothermal and irreversible process, the work done \( W \) is given by:
\[ W = -P_{\text{external}} (V_2 - V_1) \]
Since we are assuming \( P_{\text{external}} = P_2 \):
\[ W = -P_2 (V_2 - V_1) \]
Substituting the values:
\[ W = -99600 \, \text{Pa} \times (25 \times 10^{-3} - 5 \times 10^{-3}) \]
\[ W = -99600 \times (0.025 - 0.005) \]
\[ W = -99600 \times 0.020 \]
\[ W = -1992 \, \text{J} \]

### Step 5: Convert the work done to kilojoules
Since 1 kJ = 1000 J:
\[ W = -1.992 \, \text{kJ} \approx -1.99 \, \text{kJ} \]

### Final Answer
The work done in the process is approximately:
\[ W \approx -1.99 \, \text{kJ} \]