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First law of thermodynamics. Gas expansion
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[QUOTE="Sweden08, post: 1894961, member: 146302"] [b]1.1 kg off air with pressure 10^6 Pa, temperature 125 celciusdegrees, gets to expand until the volume is five times bigger than in the beginning. In every moment of the expansion, the amount of (warmth energy = Q) contributed is one fourth of the work done by the gas. 1 kmol of air weighs about 29 kg, Cv=5/2R. Calculate the pressure after the expansion.[/b] [b]2. dU = dQ - dW dU = dQ - pdV U = 5/2 NKT[/b] [b]3. Since this process is not konstant for either temperature, volume, pressure, i have no clue how to solve it. However: If dQ = 1/4 dW, then dU = 1/4 dW - dW = -3/4 dW And since U = 5/2 NKT = 5/2 PV. ---> dU = 5/2 (P1V1-P0V0) P0= start pressure, V0=start volume, V1= endvolume = 5V0, P1=endpressure=?[/b] [/QUOTE]
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First law of thermodynamics. Gas expansion
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