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A quantity of 0.50 mole of an ideal gas at 20°C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 5.0 L. Entropy change of the system is:
- a)6.7 JK–1
- b)-2.8 JK–1
- c)3.9 JK–1
- d)None is correct
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A quantity of 0.50 mole of an ideal gas at 20°C expands isothermal...
Understanding the Problem
To determine the entropy change of the system, we analyze the isothermal expansion of an ideal gas. The given parameters are:
- Moles of gas (n) = 0.50 moles
- Initial volume (V1) = 1.0 L
- Final volume (V2) = 5.0 L
- Temperature (T) = 20°C = 293 K
- Constant external pressure (P_ext) = 2.0 atm
Entropy Change Formula
The change in entropy (\(\Delta S\)) for an ideal gas during an isothermal process can be calculated using the formula:
\[
\Delta S = nR \ln \left(\frac{V_2}{V_1}\right)
\]
where:
- \(R\) = 8.314 J/(mol·K)
Calculating the Volumes Ratio
- \(\frac{V_2}{V_1} = \frac{5.0 \, \text{L}}{1.0 \, \text{L}} = 5\)
Substituting Values
Now, substituting the values into the entropy change formula:
\[
\Delta S = 0.50 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times \ln(5)
\]
Calculating \(\ln(5)\):
- \(\ln(5) \approx 1.609\)
Final Calculation
Now, substituting \(\ln(5)\):
\[
\Delta S = 0.50 \times 8.314 \times 1.609 \approx 6.7 \, \text{J/K}
\]
Conclusion
The entropy change of the system during the isothermal expansion is:
- Answer: 6.7 J/K
Thus, option 'A' is indeed correct.
To determine the entropy change of the system, we analyze the isothermal expansion of an ideal gas. The given parameters are:
- Moles of gas (n) = 0.50 moles
- Initial volume (V1) = 1.0 L
- Final volume (V2) = 5.0 L
- Temperature (T) = 20°C = 293 K
- Constant external pressure (P_ext) = 2.0 atm
Entropy Change Formula
The change in entropy (\(\Delta S\)) for an ideal gas during an isothermal process can be calculated using the formula:
\[
\Delta S = nR \ln \left(\frac{V_2}{V_1}\right)
\]
where:
- \(R\) = 8.314 J/(mol·K)
Calculating the Volumes Ratio
- \(\frac{V_2}{V_1} = \frac{5.0 \, \text{L}}{1.0 \, \text{L}} = 5\)
Substituting Values
Now, substituting the values into the entropy change formula:
\[
\Delta S = 0.50 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times \ln(5)
\]
Calculating \(\ln(5)\):
- \(\ln(5) \approx 1.609\)
Final Calculation
Now, substituting \(\ln(5)\):
\[
\Delta S = 0.50 \times 8.314 \times 1.609 \approx 6.7 \, \text{J/K}
\]
Conclusion
The entropy change of the system during the isothermal expansion is:
- Answer: 6.7 J/K
Thus, option 'A' is indeed correct.
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A quantity of 0.50 mole of an ideal gas at 20°C expands isothermal...
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A quantity of 0.50 mole of an ideal gas at 20°C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 5.0 L. Entropy change of the system is:a)6.7 JK–1b)-2.8 JK–1c)3.9 JK–1d)None is correctCorrect answer is option 'A'. Can you explain this answer?
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