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  • One mole of a monatomic ideal gas is taken through the cycle shown in figure. <img alt="\mathrm{ A \rightarrow B }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20A%2

One mole of a monatomic ideal gas is taken through the cycle shown in figure.

\mathrm{ A \rightarrow B } Adiabatic expansion

\mathrm{ B \rightarrow C } Cooling at constant volume

\mathrm{ C \rightarrow D } Adiabatic compression.

\mathrm{ D \rightarrow A } Heating at constant volume

The pressure and temperature at \mathrm{ A,B } etc,. are denoted by \mathrm{ P_A, T_A ; P_B, T_B} etc. respectively. Given \mathrm{T_A=1000 K, P_B=(2 / 3) P_A \: and \: P_C=(1 / 3) P_A}. The work done by the gas in the process \mathrm{A \rightarrow B}  is


 

Option: 1

\mathrm{150R}


Option: 2

\mathrm{200R}


Option: 3

\mathrm{225R}


Option: 4

\mathrm{250R}


Answers (1)

best_answer

As for adiabatic change \mathrm{\mathrm{PV}^{\mathrm{\gamma}}= constant}

\mathrm{ \text { i.e. } P\left(\frac{\mu R T}{P}\right)^\gamma=\text { cons } \tan t[\operatorname{asPV}=\mu \mathrm{RT}] }

\mathrm{ i.e. \frac{\mathrm{T}^{\mathrm{\gamma }}}{\mathrm{P}^{\mathrm{\gamma }}-1}= constant\: so \: \left(\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}\right)^{\mathrm{Y}}=\left(\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}\right)^{\mathrm{Y}-1} with \mathrm{Y}=\frac{5}{3} }

\mathrm{ i.e. T_B=T_A\left(\frac{2}{3}\right)^{1-\frac{1}{\gamma }}=1000\left(\frac{2}{3}\right)^{2 / 5}=850 \mathrm{~K} }

\mathrm{ so \: \mathrm{W}_{\mathrm{AB}}=\frac{\mu \mathrm{R}\left[\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right]}{[1-\mathrm{\gamma }]}=\frac{1 \times 8.31[1000-850]}{[(5 / 3)-1]} }

\mathrm{i.e. \: W_{A B}=(3 / 2) \times R \times 150=225 R}




 

Posted by

Pankaj Sanodiya

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