Question

The work done during reversible isothermal expansion of one mole of hydrogen gas at 25^oC from a pressure of 20 atm to 10 atm is:
(Given R = 2.0 cal K^-1 mol^-1)

• A. 0 calorie
• B. -413.14 calories
• C. 413.14 calories
• D. 100 calories

Answer 1

The Correct Option is B

1. For isothermal expansion, work done is calculated as:

\(W = -nRT \ln \frac{P_2}{P_1}\)

 

2. Substituting the given values:

 

\(W = -(1) \times (2.0) \times (298) \times \ln \left(\frac{10}{20}\right)\)

 

3. Simplify:

 

\(W = -596 \times \ln(0.5) = -596 \times (-0.693) \approx -413.14 \text{ calories}\).

Answer 2

Step 1: Use the Formula for Work Done During Isothermal Expansion

The work done during isothermal expansion is given by:

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

• n = 1 mol (number of moles)
• R = 2.0 cal mol⁻¹K⁻¹ (gas constant)
• T = 25°C = 298 K (temperature)
• P₁ = 20 atm
• P₂ = 10 atm

Step 2: Substitute Values into the Equation

Substitute the given values into the equation:

$$ W = -1 \cdot 2.0 \cdot 298 \cdot \ln \left( \frac{10}{20} \right) $$

Simplifying:

$$ W = -2.0 \cdot 298 \cdot \ln \left( \frac{1}{2} \right) $$

Since \( \ln \left( \frac{1}{2} \right) = -0.693 \), we get:

$$ W = -2.0 \cdot 298 \cdot (-0.693) = 413.14 \, \text{calories} $$

Step 3: Conclusion

The correct answer is:

Option (2): −413.14 calories.