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  • A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratios are <img alt="\mathrm{\frac{V_{B}}{V_{A}}=2}" src="https://entranceco

A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratios are \mathrm{\frac{V_{B}}{V_{A}}=2} and \mathrm{\frac{V_{D}}{V_{A}}=4}. If the temperture \mathrm{\mathrm{T}_{\mathrm{A}}} at \mathrm{\mathrm{A}} is \mathrm{ 27 { }^{\circ} \mathrm{C}}.  Then the temperature of the gas at point \mathrm{ \mathrm{B}} is:

Option: 1

600 \mathrm{~K}


Option: 2

800 \mathrm{~K}


Option: 3

950 \mathrm{~K}


Option: 4

1150 \mathrm{~K}


Answers (1)

best_answer

Given :
Number of moles, \mathrm{n}=2
\mathrm{\mathrm{T}_{\mathrm{A}}=27^{\circ} \mathrm{C}=300 \mathrm{~K}}

\mathrm{Let \, \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{0} then \mathrm{V}_{\mathrm{B}}=2 \mathrm{~V}_{0} and \mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}=4 \mathrm{~V}_{0}}


\mathrm{Process \, A \rightarrow B}:

\mathrm{\mathrm{V} \propto \mathrm{T} \quad \Rightarrow \frac{T_{B}}{T_{A}}=\frac{V_{B}}{V_{A}}}
\mathrm{\therefore \mathrm{T}_{\mathrm{B}}=\mathrm{T}_{\mathrm{A}}\left(\frac{V_{B}}{V_{A}}\right)=(300)(2)=600 \mathrm{~K} }
\mathrm{\therefore \mathrm{T}_{\mathrm{B}}=600 \mathrm{~K} }

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