Two moles of helium gas $γ = 5 / 3$ at $27^{\circ} C$ is expanded at constant pressures until its volume is doubled. Then it undergoes an adiabatic change until the temperatures returns to its initial value. The work done during adiabatic process is _____ (universal gas constant = $8.3 J m o l^{- 1} K^{- 1}$ )

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7470J

4780J

4770J

7070J

answer is A.

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Detailed Solution

Helium gas constant $γ = 5 / 3$

Temperature, $T = 27 º C$

$\Rightarrow T = 273 + 27 = 300 K$

$\begin{array}{l} Initial Volume, V_{1} = V, \\ {Final volumeV}_{2} = 2V, \end{array}$

$At constant pressure, P = constant,$

$V_{1} / T_{1} = V_{2} / T_{2}, T_{2} = {2T}_{1} {,T}_{2} = 600 K$

$\begin{array}{l} Work done, \\ W = {nR (T_{2} - T_{1})} / (γ - 1), = (\frac{2 \times 8.3 \times 300}{\frac{2}{3}}) = 7470 Joule \end{array}$

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