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  • Calculate the change in entropy when 5 moles of an ideal gas expands reversibly and isothermally from an initial volume of 20 liters to 100 liters at <img alt="400 \mathrm{~K}." src="https://entran

Calculate the change in entropy when 5 moles of an ideal gas expands reversibly and isothermally from an initial volume of 20 liters to 100 liters at 400 \mathrm{~K}.
 

Option: 1

66.9 \mathrm{~J} / \mathrm{K}

 


Option: 2

69 \mathrm{~J} / \mathrm{K}
 


Option: 3

71 \mathrm{~J} / \mathrm{K}
 


Option: 4

75 \mathrm{~J} / \mathrm{K}


Answers (1)

best_answer

Step 1: Use the ideal gas law to relate the initial and final conditions.

\mathrm{ P V=n R T }

Step 2: Calculate the initial and final pressures using the ideal gas law.

\mathrm{ P_1=\frac{n R T_1}{V_1} }

\mathrm{ P_2=\frac{n R T_2}{V_2} }

Step 3: Use the isothermal process to find the change in entropy.

\mathrm{ \Delta S=n R \ln \left(\frac{V_2}{V_1}\right) }

Step 4: Substitute the given values and calculate the change in entropy.

\mathrm{ \Delta S=(5 \mathrm{~mol}) \times(8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}) \times \ln \left(\frac{100 \mathrm{~L}}{20 \mathrm{~L}}\right) }

Calculating the numerical value of the change in entropy:

\mathrm{ \Delta S \approx 5 \times 8.314 \times \ln (5) \mathrm{J} / \mathrm{K} }

\mathrm{ \Delta S=66.9 \mathrm{~J} / \mathrm{K} }

So, correct option is 1.

Posted by

SANGALDEEP SINGH

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