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An ideal gas is taken through the cycle \mathrm{A \rightarrow B \rightarrow C \rightarrow A}, as shown in the figure. If the net heat supplied to the gas in the cycle is \mathrm{5 \mathrm{~J}}, the work done by the gas in the process \mathrm{C \rightarrow A} is

Option: 1

-5 \mathrm{~J}


 


Option: 2

-10 \mathrm{~J}
 


Option: 3

-15 \mathrm{~J}
 


Option: 4

-{2 0} \: \mathrm{J}


Answers (1)

best_answer

For the cyclic process \mathrm{\Delta \mathrm{U}=0}

\mathrm{ \Delta \mathrm{W}=W_{A B}+W_{B C}+W_{C A}=\left(10+0+W_{C A}\right) \mathrm{J} }

Given \mathrm{ \Delta Q=5 \mathrm{~J} }

From first law of thermodynamics \mathrm{ 5=10+0+W_{C A} }

\mathrm{ \Rightarrow W_{C A}=-5 }

Hence option 1 is correct.



 

Posted by

sudhir kumar

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