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A monatomic ideal gas of two moles is taken through a cyclic process starting for \mathrm{A} as shown in the figure. The volume ratios are \mathrm{\left(V_B / V_A\right)=2 \: and\: \left(V_D / V_A\right)=4}. If the temperature \mathrm{T_A \: at \: A\: is \: 27^{\circ} \mathrm{C}}    . The temperature of the gas at point \mathrm{B} is


 

Option: 1

27^{\circ} \mathrm{C}


 


Option: 2

127^{\circ} \mathrm{C}
 


Option: 3

227^{\circ} \mathrm{C}
 


Option: 4

327^{\circ} \mathrm{C}


Answers (1)

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Taking \mathrm{V}_{\mathrm{A}}=\mathrm{V}_0 ; \quad \mathrm{V}_{\mathrm{B}}=2 \mathrm{~V}_0 ; \mathrm{V}_{\mathrm{D}}=4 \mathrm{~V}_0

Process \mathrm{\mathrm{A} \rightarrow \mathrm{B}} : Isobaric process as \mathrm{\mathrm{V} / \mathrm{T}=} constant

\mathrm{ \mathrm{V}_0 /(273+27)=\left(2 \mathrm{~V}_0\right) / \mathrm{T}_{\mathrm{B}} }

\mathrm{ \Rightarrow \mathrm{T}_{\mathrm{B}}=600 \mathrm{~K}=327^{\circ} \mathrm{C} \text {. } }

 

Posted by

Divya Prakash Singh

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