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# Engineering

Following figure shows two processes A and B for a gas. If \Delta Q_{A} and \Delta Q_{B} are the amount of heat absorbed by the system in two cases, and 

\Delta U_{A} and \Delta U_{B} are changes in internal energies, respectively, then :

 

  • Option 1)

     \Delta Q_{A}< \Delta Q_{B},\Delta U_{A}< \Delta U_{B}   

  • Option 2)

    \Delta Q_{A}> \Delta Q_{B},\Delta U_{A}> \Delta U_{B}

  • Option 3)

    \Delta Q_{A}> \Delta Q_{B},\Delta U_{A}= \Delta U_{B}

  • Option 4)

     \Delta Q_{A}= \Delta Q_{B},\Delta U_{A}= \Delta U_{B}

 

Answers (1)
S solutionqc

Internal energy is state function

So,     u_{A}=u_{B}

Now,    \Delta Q=\Delta u- w

Now    w_{A}> w_{B}  (work is Area under the curve

So,     \Delta Q_{A}> Q_{B}


Option 1)

 \Delta Q_{A}< \Delta Q_{B},\Delta U_{A}< \Delta U_{B}   

Option 2)

\Delta Q_{A}> \Delta Q_{B},\Delta U_{A}> \Delta U_{B}

Option 3)

\Delta Q_{A}> \Delta Q_{B},\Delta U_{A}= \Delta U_{B}

Option 4)

 \Delta Q_{A}= \Delta Q_{B},\Delta U_{A}= \Delta U_{B}

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