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Helium gas goes through a cycle ABCDA ( consisting of two isochoric and two isobaric lines ) as shown in figure. Efficiency of this cycle is nearly: ( Assume the gas to be close to ideal gas )

Option: 1

15.4


Option: 2

9.1


Option: 3

10.5


Option: 4

12.5


Answers (1)

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In the case of a cyclic process, work done is equal to the area under the cycle and is taken to be positive if the cycle is clockwise.

\therefore    Work is done by the gas

        W= Area of the rectangle ABCD=P0V0

\therefore    Helium gas is a monoatomic gas

\therefore \: \: C_{V}= \frac{3}{2}R\: and \: C_{p}= \frac{5}{2}R

Along the path AB, heat supplied to the gas at  constant volume,

\therefore\: \: \: \Delta Q_{AB}= nC_{V}\Delta T= n\frac{3}{2}R\Delta T= \frac{3}{2}V_{0}\Delta P= \frac{3}{2}P_{0}V_{0}

Along the path BC, heat supplied to the gas at  constant pressure,

\therefore\: \: \: \Delta Q_{BC}= nC_{P}\Delta T= n\frac{ 5}{2}R\Delta T= \frac{5}{2}\left ( 2P_{0} \right )\Delta V=5P_{0}V_{0}

Along the path CD and DA, heat is rejected by the gas 

\mathrm{Efficiency\: ,\eta = \frac{work \: done\: by\: the \: gas}{Heat \: supply\: to\: the \: gas }\times 100}

\mathrm{= \frac{P_{0}V_{0}}{\frac{3}{2}P_{0}V_{0}+5P_{0}V_{0}}\times 100= \frac{200}{13}= 15.4 \%}

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Suraj Bhandari

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