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sleet_krkn [62]
3 years ago
6

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperatu

re of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you fill in the following table. P V T A 1 400 kPa 10.0 L 720 K B C 24 L D 15 L (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A ? B, B ? C, C ? D, and D ? A. Process Q (kJ) W (kJ) ?Eint (kJ) A ? B B ? C C ? D D ? A (c) Calculate the efficiency Wnet / |Qh|. % (d) Show that the efficiency is equal to 1 ? TC / TA, the Carnot efficiency. (Do this on paper. Your instructor may ask you to turn in this work.)

Physics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

Find attached the solution

Explanation:

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Answer:

The current is reduced to half of its original value.

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        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

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