D Zwischenergebnis: 0=S₂²+ (Po-S₁)S₂-P151 A Pe 1 bar Wasserbad 0 = 50°C Abbildung 5 Universelle Gaskonstante: Gas P-5 ba

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answerhappygod
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D Zwischenergebnis: 0=S₂²+ (Po-S₁)S₂-P151 A Pe 1 bar Wasserbad 0 = 50°C Abbildung 5 Universelle Gaskonstante: Gas P-5 ba

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D Zwischenergebnis 0 S Po S S P151 A Pe 1 Bar Wasserbad 0 50 C Abbildung 5 Universelle Gaskonstante Gas P 5 Ba 1
D Zwischenergebnis 0 S Po S S P151 A Pe 1 Bar Wasserbad 0 50 C Abbildung 5 Universelle Gaskonstante Gas P 5 Ba 1 (15.36 KiB) Viewed 42 times
In a cylindrical container with a base area of ​​𝐴 = 0.5 𝑚2 there is an ideal gas at a pressure of 𝑝1 = 5 𝑏𝑎𝑟. The ambient pressure is 𝑝∞ = 1 𝑏𝑎𝑟. The piston is initially fixed at a distance from the tank bottom of 𝑠1 = 0.4 𝑚. In this state, the spring attached to the piston (see Figure 5) is relieved. If the spring is compressed, an additional
force on the piston. The spring constant is 𝐷 = 500 𝑘𝑁. Then the piston is released and 𝑚
slowly into the new position of rest (balance of forces, the weight of the piston can be neglected). Calculate the pressure of the gas for this condition, assuming that the piston is moved without dissipation and the gas in the cylinder is kept constant at 50 °C by the surrounding water bath. What work did the gas do and what amount of heat was exchanged between the water bath and the gas in the cylinder? What is the amount of substance of the gas in the container?
D Zwischenergebnis 0 S Po S S P151 A Pe 1 Bar Wasserbad 0 50 C Abbildung 5 Universelle Gaskonstante Gas P 5 Ba 2
D Zwischenergebnis 0 S Po S S P151 A Pe 1 Bar Wasserbad 0 50 C Abbildung 5 Universelle Gaskonstante Gas P 5 Ba 2 (15.36 KiB) Viewed 42 times
D Zwischenergebnis: 0=S₂²+ (Po-S₁)S₂-P151 A Pe 1 bar Wasserbad 0 = 50°C Abbildung 5 Universelle Gaskonstante: Gas P-5 bar R = 8,314 mol K Feder Kolben
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