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5 moles of Hydrogen\left(\gamma=\frac{7}{5}\right) initially at S.T.P. are compressed adiabatically so that its temperature becomes 400^{\circ} \mathrm{C}. The increase in the internal energy of the gas in kilo-joules is:

\left(R=8.30 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)

Option: 1

21.55


Option: 2

41.50


Option: 3

65.55


Option: 4

80.55


Answers (1)

best_answer

\mathrm{\text { Given } T_1=0^{\circ} \mathrm{C}=273 \mathrm{~K}, T_2=400^{\circ} \mathrm{C}=673 \mathrm{~K}}

\mathrm{\begin{aligned} & \text { Work done } W=\frac{n R}{(\gamma-1)}\left(T_2-T_1\right)=\frac{5 \times 8.3 \times 400}{\left(\frac{7}{5}-1\right)} \\ & =41500 \mathrm{~J}=41.5 \mathrm{~kJ} \end{aligned}}

By convention, the work done on the gas is taken to be negative, i.e. W = −41.5 kJ From
the first law of thermodynamics dQ = dU + dW For an adiabatic process, dQ = 0
Hence dU = −dW = −(−41.5) = 41.5 kJ The positive sign of dU implies that the
internal energy increases

Posted by

Ajit Kumar Dubey

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