Calculate the work done when 1 mole of a gas expands reversibly and isothermally from 5 atm to 1 atm at 300 K. [Value of log 5 = 0.6989].

Options

A. `-4015` J
B. `+4015` J
C. Zero
D. `-15` J

Answer

To calculate the work done when 1 mole of a gas expands reversibly and isothermally from 5 atm to 1 atm at 300 K, we can use the formula for work done during an isothermal expansion of an ideal gas:

\[ W = -nRT \ln\left(\frac{P_2}{P_1}\right) \]

Where:
- \( W \) = work done (in Joules)
- \( n \) = number of moles of gas (1 mole)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (300 K)
- \( P_1 \) = initial pressure (5 atm)
- \( P_2 \) = final pressure (1 atm)

### Step-by-step Solution:

1. **Identify the values:**
- \( n = 1 \) mole
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 300 \, \text{K} \)
- \( P_1 = 5 \, \text{atm} \)
- \( P_2 = 1 \, \text{atm} \)

2. **Calculate the ratio of pressures:**
\[ \frac{P_2}{P_1} = \frac{1 \, \text{atm}}{5 \, \text{atm}} = 0.2 \]

3. **Calculate the natural logarithm of the pressure ratio:**
\[ \ln(0.2) = \ln\left(\frac{1}{5}\right) = -\ln(5) \quad \text{(Using properties of logarithms)} \]
Given \( \log(5) = 0.6989 \), we can find:
\[ \ln(5) = 2.303 \times \log(5) = 2.303 \times 0.6989 \approx 1.607 \]
Therefore,
\[ \ln(0.2) \approx -1.607 \]

4. **Substitute the values into the work done formula:**
\[ W = -nRT \ln\left(\frac{P_2}{P_1}\right) \]
\[ W = -1 \times 0.0821 \times 300 \times (-1.607) \]

5. **Calculate the work done:**
\[ W = 0.0821 \times 300 \times 1.607 \]
\[ W \approx 39.63 \, \text{L·atm} \]

6. **Convert L·atm to Joules:**
\[ 1 \, \text{L·atm} = 101.325 \, \text{J} \]
\[ W \approx 39.63 \times 101.325 \approx 4015.5 \, \text{J} \]

7. **Final answer with sign:**
Since the gas is expanding, the work done by the system is negative:
\[ W \approx -4015.5 \, \text{J} \]

### Final Result:
The work done when 1 mole of gas expands is approximately **-4015.5 J**.