Question

23. On expanding 5 moles of H2 gas at NTP its temperature decreases to 260 K. Find the work done by the gas. Find the change in internal energy of the gas. (Given y = 1.4 and R = 8.31 J/mol-K) Ans.1350 J, decreases by 1350 J

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Answer 1

Answer:

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.