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general chemistry4 - 5.60 Spring 2004 Lecture#6 page 1...

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5.60 Spring 2004 Lecture #6 page 1 Reversible Adiabatic Expansion (or compression) of an Ideal Gas one mole ideal gas reversible adiabatic 1 mol gas ( V 1 , T 1 ) = 1 mol gas ( V 2 , T 2 ) Adiabatic dq = 0 Reversible dw = -pdV Ideal gas dU = C v dT \ From 1 st Law dU = -pdV C v dT = -pdV along path C V dT = - pdV p = RT V C V dT T = - R dV V Integrating: C V dT T T 1 T 2 = - R dV V V 1 V 2 C V ln T 2 T 1 = - R ln V 2 V 1 T 2 T 1 C V = V 1 V 2 R T 2 T 1 = V 1 V 2 R C V C p - C V = R for i.g. æ ææææ æ T 2 T 1 = V 1 V 2 C p C V - 1 Define g C p C V T 2 T 1 = V 1 V 2 g - 1 For monatomic ideal gas: C V = 3 2 R C p = 5 2 R g = 5 3 ( > 1 generally) Since g > 1 g - 1 > 0
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5.60 Spring 2004 Lecture #6 page 2 In an adiabatic expansion ( V 2 > V 1 ), the gas cools ( T 2 < T 1 ). And in an adiabatic compression ( V 2 < V 1 ), the gas heats up. For ideal gas (one mole) T = pV R p 2 p 1 = V 1 V 2 g In other words p 1 V 1 g = p 2 V 2 g pV g is constant along a reversible adiabat For isothermal process T = constant pV = constant pV is constant along an isotherm V 2 adiabat < V 2 isotherm because the gas
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