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Two moles of Helium gas \mathrm{(\gamma=5 / 3)} are initially at temperature \mathrm{27^{\circ} \mathrm{C}} and occupy a volume of 20 litres. The gas is first expanded at the constant pressure until volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.What is the work done by the gas? \mathrm{\text{(Gas constant R = 8.3 T/mole K)}}

Option: 1

\mathrm{13450\: J}


Option: 2

\mathrm{14450\: J}


Option: 3

\mathrm{16450\: J}


Option: 4

\mathrm{12450\: J}


Answers (1)

best_answer

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 \mathrm{ BC }

\mathrm{ \mathrm{W}_2=\frac{\mathrm{nR}}{1-\gamma}\left[\mathrm{T}_2-\mathrm{T}_1\right]=\frac{2 \times 8.3}{1-\left(\frac{5}{3}\right)}[300-600]=7470 \mathrm{~J} \text {. } }

\mathrm{ \therefore \quad } Net work done by gas \mathrm{ \mathrm{W}=\mathrm{W}_1+\mathrm{W}_2 }

\mathrm{ =4980+7470=12450 \mathrm{~J} }
 

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shivangi.bhatnagar

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