A quantity of an ideal gas at `20^(@)C` reversibly expands against a constant pressure of 2.0 atm from 1.0 L to 2.0 L. Calculate the work done

Options

A. `-101.3` J
B. `-202.6` J
C. `-844` J
D. `-448`J

Answer

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

### Step 1: Identify the given values
- Initial volume (V1) = 1.0 L
- Final volume (V2) = 2.0 L
- Constant pressure (P) = 2.0 atm

### Step 2: Calculate the change in volume (ΔV)
The change in volume (ΔV) can be calculated using the formula:
\[ \Delta V = V2 - V1 \]
Substituting the values:
\[ \Delta V = 2.0 \, \text{L} - 1.0 \, \text{L} = 1.0 \, \text{L} \]

### Step 3: Calculate the work done (W)
The work done by the gas during expansion against a constant pressure is given by the formula:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = - (2.0 \, \text{atm}) \times (1.0 \, \text{L}) = -2.0 \, \text{L} \cdot \text{atm} \]

### Step 4: Convert work done from L·atm to Joules
To convert the work done from L·atm to Joules, we use the conversion factor:
\[ 1 \, \text{L} \cdot \text{atm} = 101.325 \, \text{J} \]
Thus,
\[ W = -2.0 \, \text{L} \cdot \text{atm} \times 101.325 \, \text{J/L·atm} = -202.65 \, \text{J} \]

### Step 5: Round the answer
Rounding the answer to one decimal place, we get:
\[ W \approx -202.6 \, \text{J} \]

### Final Answer
The work done during the reversible expansion of the gas is approximately:
\[ W = -202.6 \, \text{J} \]

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