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2 litres of an ideal gas : V1 = 2 litres
P1 = 10 atms
T = constant
V2 = 10 litres
   =>  P2 = P1 V1 / V2 = 2 atm

work done =
W=\int\limits^{V_2}_{V_1} {P} \, dV \\\\W=\int\limits^{V_2}_{V_1} {\frac{P_1*V_1}{V}} \, dV = 20\ [Ln\ V]_{V_1}^{V_2}\\\\=20*[Ln\ \frac{V_2}{V_1}]\ atm-litres\\\\=20*Ln_e\ 5\ *1.013*10^5\ Pa*10^{-3}m^3\\=3260.7 Joules

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in the internal energy.

So energy absorbed is = 3.2607 kJoules.

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