Two moles of an ideal gas at 2 atm and `27^(@)C ` is compressed isothermally to half of its volume by external pressure of 4 atm. The work doen is

Options

A. 4.985kJ
B. 49.2kJ
C. 98.4 kJ
D. 101.32 kJ

Explanation

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

### Step 1: Understand the Given Information
- Number of moles (n) = 2 moles
- Initial pressure (P1) = 2 atm
- Final pressure (P2) = 4 atm (external pressure)
- Initial temperature (T) = 27°C = 300 K (convert to Kelvin)
- The gas is compressed isothermally to half its volume.

### Step 2: Calculate Initial Volume (V1)
Using the ideal gas law, \( PV = nRT \):
\[ V_1 = \frac{nRT}{P_1} \]
Where:
- R = 0.0821 L·atm/(K·mol) (ideal gas constant)

Substituting the values:
\[ V_1 = \frac{2 \text{ moles} \times 0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}}{2 \text{ atm}} \]
\[ V_1 = \frac{49.26 \text{ L·atm}}{2 \text{ atm}} = 24.63 \text{ L} \]

### Step 3: Calculate Final Volume (V2)
Since the gas is compressed to half its volume:
\[ V_2 = \frac{V_1}{2} = \frac{24.63 \text{ L}}{2} = 12.315 \text{ L} \]

### Step 4: Calculate Work Done (W)
The work done on the gas during isothermal compression can be calculated using the formula:
\[ W = -P_{ext} \Delta V \]
Where:
- \( \Delta V = V_2 - V_1 \)

Calculating \( \Delta V \):
\[ \Delta V = 12.315 \text{ L} - 24.63 \text{ L} = -12.315 \text{ L} \]

Now substituting into the work formula:
\[ W = -4 \text{ atm} \times (-12.315 \text{ L}) \]
\[ W = 49.26 \text{ L·atm} \]

### Step 5: Convert Work to Joules
To convert L·atm to Joules, use the conversion factor \( 1 \text{ L·atm} = 101.325 \text{ J} \):
\[ W = 49.26 \text{ L·atm} \times 101.325 \text{ J/L·atm} \]
\[ W \approx 4992.5 \text{ J} \]

### Final Answer
The work done during the isothermal compression is approximately **4992.5 J**.

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