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8 kg of air undergoes a reversible adiabatic process from 2 bar, 400C to 10 bar. Find a) work transfer b) change in internal energy and c) heat transfer in the process. |
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Answer» P1 = 2 bar ⇒ 2 x 105 pascal P2 = 10 bar = 10 x 105 pascal T1 = 400 i T1 = 673 k T2 = ? First law of thermodynamics \(\Delta Q=\Delta U + \Delta W\) For adiabatic process \(\Delta Q = 0\) Then work done \(\Delta W = \Delta U\) \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) \(\frac{2\times10}{673}=\frac{10\times10}{T_2}\) T2 = 5 x 673 T2 = 3365 k (a) Work done \(\Delta W=\frac{rR}{r-1}(T_2-T_1)\) \(\Delta W=\frac{1.4\times8.3}{1.4-1}(3365-673)\) \(\Delta W=\frac{11.62}{0.4}(2692)\) \(\Delta W = 78.20 KJ\) (b) Change in internal energy is equal to work done \(\Delta U=\Delta W\) \(\Delta U=\Delta W\) (c) Heat transfer in the adiabatic process \(\Delta Q=0\) |
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