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
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
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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).
Updated On: Jun 3, 2026
- -52.18kj
- +26.09kj
- -26.09kj
- +52.18kj
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The Correct Option is C
Solution and Explanation
Concept:
For an isothermal, reversible expansion of an ideal gas, the work done ($W$) is calculated using the formula:
\[ W = -2.303 \, nRT \log \left( \frac{P_1}{P_2} \right) \]
Where:
• $n$ is the number of moles of the gas.
• $R$ is the gas constant ($8.314 \, \text{J/mol}\cdot\text{K}$).
• $T$ is the absolute temperature.
• $P_1$ and $P_2$ are the initial and final pressures.
Step 1: Calculate the number of moles of Hydrogen ($H_2$).
Given mass = $10 \, \text{g}$. Molar mass of $H_2 = 2 \, \text{g/mol}$. \[ n = \frac{10}{2} = 5 \, \text{moles} \]
Step 2: Substitute values into the work formula.
Given $T = 273 \, \text{K}$, $P_1 = 10 \, \text{atm}$, and $P_2 = 1 \, \text{atm}$. \[ W = -2.303 \times 5 \times 8.314 \times 273 \times \log \left( \frac{10}{1} \right) \] \[ W = -2.303 \times 5 \times 8.314 \times 273 \times 1 \]
Step 3: Calculate the final value in kilojoules.
\[ W \approx -26145 \, \text{J} \] Converting to kilojoules: \[ W \approx -26.145 \, \text{kJ} \] Based on the options, the closest value is $-26.09 \, \text{kj}$.
• $n$ is the number of moles of the gas.
• $R$ is the gas constant ($8.314 \, \text{J/mol}\cdot\text{K}$).
• $T$ is the absolute temperature.
• $P_1$ and $P_2$ are the initial and final pressures.
Step 1: Calculate the number of moles of Hydrogen ($H_2$).
Given mass = $10 \, \text{g}$. Molar mass of $H_2 = 2 \, \text{g/mol}$. \[ n = \frac{10}{2} = 5 \, \text{moles} \]
Step 2: Substitute values into the work formula.
Given $T = 273 \, \text{K}$, $P_1 = 10 \, \text{atm}$, and $P_2 = 1 \, \text{atm}$. \[ W = -2.303 \times 5 \times 8.314 \times 273 \times \log \left( \frac{10}{1} \right) \] \[ W = -2.303 \times 5 \times 8.314 \times 273 \times 1 \]
Step 3: Calculate the final value in kilojoules.
\[ W \approx -26145 \, \text{J} \] Converting to kilojoules: \[ W \approx -26.145 \, \text{kJ} \] Based on the options, the closest value is $-26.09 \, \text{kj}$.
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