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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{\mathrm{D} \rightarrow \mathrm{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 temperature \mathrm{T_D} is:
\mathrm{(2 / 3)^{2 / 5}=0.85\: and\: \mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}.}


 

Option: 1

450 \mathrm{~K}
 


Option: 2

500 \mathrm{~K}
 


Option: 3

550 \mathrm{~K}
 


Option: 4

600 \mathrm{~K}


Answers (1)

best_answer

As \mathrm{A\: and \: D} are on the same isochor

\mathrm{ \frac{P_D}{P_A}=\frac{T_D}{T_A} \text {, i. e. , } P_D=P_A \frac{T_D}{T_A} }

But \mathrm{ C\: and \: D } are on the same adiabatic

\mathrm{ \left(\frac{T_D}{T_C}\right)^\gamma =\left(\frac{P_D}{P_C}\right)^{\gamma -1}=\left(\frac{P_A T_D}{P_C T_A}\right)^{\gamma -1} }

\mathrm{or \left(\mathrm{T}_{\mathrm{D}}\right)^{1 / \mathrm{\gamma }}=\mathrm{T}_{\mathrm{C}}\left[\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{C}} \mathrm{T}_{\mathrm{A}}}\right]^{1-\frac{1}{\mathrm{~\gamma }}}, \quad i.e. \mathrm{T}_{\mathrm{D}}^{3 / 5}=\left(\frac{\mathrm{T}_{\mathrm{B}}}{2}\right)\left[\frac{\mathrm{P}_{\mathrm{A}}}{(1 / 3) \mathrm{P}_{\mathrm{A}} 1000}\right]^{2 / 5} }

\mathrm{i.e. \quad \mathrm{T}_{\mathrm{D}}^{3 / 5}=\left[\frac{1}{2}\left(\frac{2}{3}\right)^{2 . / 5} \times 1000\right]\left[\frac{3}{1000}\right]^{2 / 5} i. e., \mathrm{T}_{\mathrm{D}}=500 \mathrm{~K}}




 

Posted by

Ajit Kumar Dubey

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