n moles of an ideal gas with constant volume heat capacity C_{V}

undergo an isobaric expansion by certain volume. The ratio of the work done

in the process, to the heat supplied is :

  • Option 1)

    \frac{nR}{C_{V}+nR}

  • Option 2)

    \frac{nR}{C_{V}-nR}

  • Option 3)

    \frac{4nR}{C_{V}-nR}

  • Option 4)

    \frac{4nR}{C_{V}+nR}

 

Answers (1)

 

First law in isobaric process -

\Delta U= n\, C_{v}\Delta T

= n\frac{R}{\gamma -1}\Delta T
 

- wherein

\Delta Q= \Delta U+W

         = n\frac{\gamma \: R}{\gamma -1}\cdot \Delta T

\Delta Q= nC_{p}\: \Delta T

 

 

C_V=n\frac{f}{2}R

W=nR\Delta T

\Delta Q=nC_P\Delta T=(\frac{f}{2}+1)nR\Delta T

\frac{W}{\Delta Q}=\frac{nR\Delta T}{nC_P\Delta T}=\frac{1}{\frac{f}{2}+1}=\frac{2}{2+f}

\frac{W}{\Delta Q}=\frac{2}{2+\frac{2C_V}{nR}}=\frac{2nR}{2(nR+C_V)}

\frac{W}{\Delta Q}=\frac{nR}{(nR+C_V)}


Option 1)

\frac{nR}{C_{V}+nR}

Option 2)

\frac{nR}{C_{V}-nR}

Option 3)

\frac{4nR}{C_{V}-nR}

Option 4)

\frac{4nR}{C_{V}+nR}

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