Question:
The workdone by a gas of one mole at constant temperature of $27^\circ\text{C}$ when its volume doubled is (Take $\log_{10} 2 = 0.3010$ and Universal gas constant $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$):
The workdone by a gas of one mole at constant temperature of $27^\circ\text{C}$ when its volume doubled is (Take $\log_{10} 2 = 0.3010$ and Universal gas constant $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$):
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For isothermal expansion, remember that $\ln 2 \approx 0.693$.
The work formula simplifies to $W \approx 0.693 \times R T$.
Here, $R T = 8.314 \times 300 = 2494.2\text{ J}$.
Multiplying $2494.2 \times 0.693 \approx 1728.5\text{ J}$, which immediately points to $1729\text{ J}$.
The work formula simplifies to $W \approx 0.693 \times R T$.
Here, $R T = 8.314 \times 300 = 2494.2\text{ J}$.
Multiplying $2494.2 \times 0.693 \approx 1728.5\text{ J}$, which immediately points to $1729\text{ J}$.
Updated On: Jul 22, 2026
- $1059\text{ J}$
- $1729\text{ J}$
- $1679\text{ J}$
- $865\text{ J}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We need to calculate the work done during the isothermal expansion of one mole of an ideal gas when its volume is doubled at a temperature of $27^\circ\text{C}$.
Step 2: Key Formula and Approach:
The work done $W$ during an isothermal process is given by the formula:
\[ W = n R T \ln \left( \frac{V_2}{V_1} \right) \] Converting the natural logarithm ($\ln$) to a common logarithm ($\log_{10}$):
\[ W = 2.303 \times n R T \log_{10} \left( \frac{V_2}{V_1} \right) \] where $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the absolute temperature, and $\frac{V_2}{V_1}$ is the volume expansion ratio.
Step 3: Detailed Explanation:
• Identify the given values:
Number of moles $n = 1\text{ mol}$
Temperature $T = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$
Universal gas constant $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$
Volume ratio $\frac{V_2}{V_1} = 2$
$\log_{10} 2 = 0.3010$
• Calculate work done ($W$):
Substitute the values into the isothermal work equation:
\[ W = 2.3026 \times 1 \times 8.314 \times 300 \times \log_{10}(2) \] \[ W = 2.3026 \times 8.314 \times 300 \times 0.3010 \] \[ W = 5743.3 \times 0.3010 \] \[ W \approx 1728.7\text{ J} \approx 1729\text{ J} \]
Step 4: Final Answer:
The work done by the gas is approximately $1729\text{ J}$, which corresponds to Option (B).
We need to calculate the work done during the isothermal expansion of one mole of an ideal gas when its volume is doubled at a temperature of $27^\circ\text{C}$.
Step 2: Key Formula and Approach:
The work done $W$ during an isothermal process is given by the formula:
\[ W = n R T \ln \left( \frac{V_2}{V_1} \right) \] Converting the natural logarithm ($\ln$) to a common logarithm ($\log_{10}$):
\[ W = 2.303 \times n R T \log_{10} \left( \frac{V_2}{V_1} \right) \] where $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the absolute temperature, and $\frac{V_2}{V_1}$ is the volume expansion ratio.
Step 3: Detailed Explanation:
• Identify the given values:
Number of moles $n = 1\text{ mol}$
Temperature $T = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$
Universal gas constant $R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$
Volume ratio $\frac{V_2}{V_1} = 2$
$\log_{10} 2 = 0.3010$
• Calculate work done ($W$):
Substitute the values into the isothermal work equation:
\[ W = 2.3026 \times 1 \times 8.314 \times 300 \times \log_{10}(2) \] \[ W = 2.3026 \times 8.314 \times 300 \times 0.3010 \] \[ W = 5743.3 \times 0.3010 \] \[ W \approx 1728.7\text{ J} \approx 1729\text{ J} \]
Step 4: Final Answer:
The work done by the gas is approximately $1729\text{ J}$, which corresponds to Option (B).
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