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  • Two moles of Helium gas are initially at temperature <img alt="27^{\circ} \mathrm{

Two moles of Helium gas (\gamma=5 / 3) are initially at temperature 27^{\circ} \mathrm{C} and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.

Question : What is the work done by the gas? (Gas constant \mathrm{R}=8.3 \mathrm{~T} / \mathrm{mole} \mathrm{K} )

Option: 1

13450 \mathrm{~J}


Option: 2

14450 \mathrm{~J}


Option: 3

16450 \mathrm{~J}


Option: 4

12450 \mathrm{~J}


Answers (1)

 The work done by gas in isobaric process \mathrm{A B}

\mathrm{=2.49 \times 10^{5} \times(40-20) \times 10^{-3}=4980 \mathrm{~J}}

The work done by gas during adiabatic process BC
\mathrm{\mathrm{W}_{2}=\frac{\mathrm{nR}}{1-\gamma}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=\frac{2 \times 8.3}{1-(5 / 3)}[300-600]=7470 \mathrm{~J}}.
\mathrm{\therefore \quad}  Net work done by gas

\mathrm{\mathrm{W}=\mathrm{W}_{1}+\mathrm{W}_{2}}
\mathrm{=4980+7470=12450 \mathrm{~J}}




 

Posted by

Sumit Saini

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