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Three moles of an ideal gas at 300 \mathrm{~K} are isothermally expanded to five times its volume and heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 83.14 \mathrm{~kJ} heat is required. Calculate the ratio \mathrm{\left(C_P / C_V\right)} of the gas.

\mathrm{\left[\log _{\mathrm{e}} 5=1.61\right. and \left.R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}^{-1}\right]}
 

Option: 1

1.2


Option: 2

1.3


Option: 3

1.4


Option: 4

1.5


Answers (1)

According to first law of thermodynamics,

\mathrm{ Q=\Delta U+W }

For an isothermal change,

\mathrm{ T=\text { constant, } U=\text { constant, } \Delta U=0 }

and \mathrm{ W=n R T \log _{\mathrm{e}}\left|\frac{V_f}{V_i}\right| }

\mathrm{ i.e. W=3 \times 8.31 \times 300 \times \log _e(5) }

\mathrm{ =12.03 \mathrm{~kJ} }

\mathrm{ \therefore Q_{\text {isothermal }}=0+12.03=12.03 \mathrm{~kJ} }            .........(i)

For isochoric change as \mathrm{ V= } constant,

\mathrm{W=\int P d V=0 }

\mathrm{\Delta U=n C_v \Delta T=3 C_v \Delta T \quad(\because \quad n=3)}

Applying gas equation between points \mathrm{A \: and \: C,}

\mathrm{\frac{P V}{300}=\frac{P(5 \mathrm{~V})}{T_C} \quad i.e. T_C=1500 \mathrm{~K} }

\mathrm{so \: that \Delta T=T_C-T_B=1500-300=1200 }

\mathrm{\therefore \quad \Delta U=3 C_v \times 1200=3.6 C_v \mathrm{~kJ}}

Note: \mathrm{T_0 \: find \: T_C} you can apply gas equation between points \mathrm{B\: and \: C} also.

and hence,
\mathrm{ (\Delta Q)_{\text {isochoric }}=3.6 C_v+0=3.6 C_v \mathrm{~kJ} }        .........(ii)

According to given problem,

\mathrm{ \Delta Q_{\text {isothermal }}+\Delta Q_{\text {isochoric }}=83.14 \mathrm{~kJ} }

Using equation (i) and (ii), we get

\mathrm{ 12.03+3.6 C_v=83.14 }

\mathrm{ \text { or } C_v=(71.11 / 3.6)=19.75 \mathrm{~J} }

\mathrm{ \text { Thus } C_P=C_V+R=19.75+8.3=28.05 \mathrm{~J} / \mathrm{mol}-\mathrm{K} }

\mathrm{\therefore \gamma=\frac{C_P}{C_V}=\frac{28.05}{19.75}=1.40}






 

 




 

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Kshitij

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