The cylinder sketched at the bottom is adiabatic on its outer walls. The piston inside the...

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The cylinder sketched at the bottom is adiabatic on its outer walls. The piston inside the cylinder seperates the two systems A and B adiabtically, the piston can move frictionless and its volume can be neglected. System B is connected by a valve to the outside. In both systems and the outside the only component is helium, that can be assumed to be a perfect gas. a) System A is compressed reversibly adiabatic by means of the piston and the rod: V42 = 0.2 VAL. The valve is open. Calculate TA2 and PA2. How much work W₁2 has to be transferred via the rod? b) Then the valve is closed and the piston rod fixed, while system B is evacuated. How big is the mass of helium, that has to be removed by the vacuum pump? c) After evacuating the whole system (consisting of subsystems A and B) it is closed to the outside by means of the valve and the pump. Due to a material defect the rod cracks resulting in a motion of the piston to the right side (irreversible process). In the final state, the piston touches the right wall. Derive a formula from the fundamental thermodynamic relation (Gibb's potential) for the change in entropy and calculate the change in entropy with it. d) After the gas from system A has expanded into the vacuum, the cylinder is heated to a temperature of 300.15 K. Calculate the change in entropy of the system. Is the result of this change of entropy AS34 valid in terms of the 2. Law? At what pressure is the system now? The following values are given: MHe = 4g.mol-¹, k = 5/3, VA VB = 1.10-3 m³, Poutside = PA = PB = 1 bar, Toutside = TA=TB = 300.15 K System A VA PA ΤΑ System B VB PB TB Figure 1: Cylinder with adiabatic outer wall. The cylinder sketched at the bottom is adiabatic on its outer walls. The piston inside the cylinder seperates the two systems A and B adiabtically, the piston can move frictionless and its volume can be neglected. System B is connected by a valve to the outside. In both systems and the outside the only component is helium, that can be assumed to be a perfect gas. a) System A is compressed reversibly adiabatic by means of the piston and the rod: V42 = 0.2 VAL. The valve is open. Calculate TA2 and PA2. How much work W₁2 has to be transferred via the rod? b) Then the valve is closed and the piston rod fixed, while system B is evacuated. How big is the mass of helium, that has to be removed by the vacuum pump? c) After evacuating the whole system (consisting of subsystems A and B) it is closed to the outside by means of the valve and the pump. Due to a material defect the rod cracks resulting in a motion of the piston to the right side (irreversible process). In the final state, the piston touches the right wall. Derive a formula from the fundamental thermodynamic relation (Gibb's potential) for the change in entropy and calculate the change in entropy with it. d) After the gas from system A has expanded into the vacuum, the cylinder is heated to a temperature of 300.15 K. Calculate the change in entropy of the system. Is the result of this change of entropy AS34 valid in terms of the 2. Law? At what pressure is the system now? The following values are given: MHe = 4g.mol-¹, k = 5/3, VA VB = 1.10-3 m³, Poutside = PA = PB = 1 bar, Toutside = TA=TB = 300.15 K System A VA PA ΤΑ System B VB PB TB Figure 1: Cylinder with adiabatic outer wall.

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a Since the compression is reversible and adiabatic the internal energy of system A does not change Therefore the change in enthalpy is zero From the ... View the full answer