Question

At 273 K the maximum work done when pressure on 10g of hydrogen is reduced from 10atm to 1 atm under isothermal, reversible conditions is

• A. -52.18kj
• B. +26.09kj
• C. -26.09kj
• D. +52.18kj

Hint:

Work done by the system (expansion) is always negative according to IUPAC convention. Since pressure is reduced (expansion), you can immediately eliminate options (b) and (d).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Work done (\(W\)) in an isothermal reversible process for an ideal gas depends on the change in volume or pressure.
Isothermal implies constant temperature. Reversible indicates maximum work.
In thermodynamics, expansion (\(P\) decreasing) means work is done by the system, which is represented with a negative sign under IUPAC convention.
Step 2: Key Formula or Approach:
\[ W = -2.303 nRT \log\left(\frac{P_1}{P_2}\right) \]
Where \(n\) is number of moles, \(R = 8.314 \text{ J/mol K}\), and \(T\) is temperature in Kelvin.
Step 3: Detailed Explanation:
1. Find the number of moles (\(n\)):
Mass = 10 g, Molar mass of \(H_2 = 2 \text{ g/mol}\).
\[ n = \frac{10}{2} = 5 \text{ moles} \]
2. Substitute values into the formula:
\(T = 273\) K, \(P_1 = 10\) atm, \(P_2 = 1\) atm.
\[ W = -2.303 \times 5 \times 8.314 \times 273 \times \log\left(\frac{10}{1}\right) \]
Since \(\log(10) = 1\):
\[ W = -2.303 \times 5 \times 8.314 \times 273 \times 1 \]
\[ W \approx -26145 \text{ J} \]
3. Convert to kilojoules:
\[ W \approx -26.145 \text{ kJ} \]
Looking at the options, the value -26.09 kJ is the closest and most accurate match.
Step 4: Final Answer:
The work done is approximately \(-26.09\) kJ.