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Entropy and Reversible Process
A certain amount of an ideal gas undergoes a reversible isothermal expansion at a temperature of 400 \mathrm{~K}. During the process, the gas absorbs 1000 \mathrm{~J} of heat from the surroundings. Calculate the change in entropy of the gas.
 

Option: 1

2.5 \mathrm{~J} / \mathrm{k}


 


Option: 2

1.5 \mathrm{~J} / \mathrm{k}
 


Option: 3

5.4 \mathrm{~J} / \mathrm{k}
 


Option: 4

2.1 \mathrm{~J} / \mathrm{k}


Answers (1)

best_answer

Given data:

Temperature, \mathrm{T=400 \mathrm{~K}}

Heat absorbed,\mathrm{ Q=1000 \mathrm{~J}}

For a reversible process, the change in entropy is given by:
\mathrm{ \Delta S=\frac{Q}{T} }
Plugging in the values:

\mathrm{ \Delta S=\frac{1000 \mathrm{~J}}{400 \mathrm{~K}} }

\mathrm{ \Delta S=2.5 \mathrm{~J} / \mathrm{K} }

Therefore, the correct option is 1

Posted by

Pankaj Sanodiya

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