The maximum work done when pressure of n moles of `H_(2)` was reduced from 20 atm to 1 atm at constant temperature of 273 K is found to be 8180 calories. What is the value of n?

Answer

To find the value of n in the given problem, we will use the formula for work done in an isothermal process. The work done (W) when the pressure of n moles of gas is reduced from an initial pressure (P_initial) to a final pressure (P_final) at constant temperature (T) is given by:

\[ W = -2.303 nRT \log \left( \frac{P_{initial}}{P_{final}} \right) \]

### Step-by-Step Solution:

1. **Identify the known values:**
- Work done (W) = -8180 calories (the negative sign indicates work done by the system)
- Initial pressure (P_initial) = 20 atm
- Final pressure (P_final) = 1 atm
- Temperature (T) = 273 K
- Gas constant (R) = 2 cal/(mol·K)

2. **Substitute the known values into the work formula:**
\[ -8180 = -2.303 \cdot n \cdot 2 \cdot 273 \cdot \log \left( \frac{20}{1} \right) \]

3. **Calculate the logarithm:**
\[ \log(20) \approx 1.301 \]

4. **Substitute the logarithm value into the equation:**
\[ -8180 = -2.303 \cdot n \cdot 2 \cdot 273 \cdot 1.301 \]

5. **Calculate the right-hand side:**
\[ -8180 = -2.303 \cdot n \cdot 2 \cdot 273 \cdot 1.301 \]
\[ -8180 = -2.303 \cdot n \cdot 2 \cdot 273 \cdot 1.301 \]
\[ -8180 = -2.303 \cdot n \cdot 2 \cdot 355.653 \]
\[ -8180 = -2.303 \cdot n \cdot 711.306 \]

6. **Remove the negative signs:**
\[ 8180 = 2.303 \cdot n \cdot 711.306 \]

7. **Solve for n:**
\[ n = \frac{8180}{2.303 \cdot 711.306} \]

8. **Calculate the value of n:**
\[ n \approx \frac{8180}{1633.36} \approx 5 \]

### Final Answer:
The value of n is 5.

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