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An engine using 1 mol of an ideal gas initially at 15.7 L and 272 K performs a cycle consisting of four steps:1) an isothermal expansion at 272 K from 15.7 L to 29.3 L; 2) cooling at constant volume to 105 K: 3) an isothermal compression to its original volume of 15.7 L; and 4) heating at constant volume to its original temperature of 272 K. Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L atm/mol/K 8.314 J/mol/K.​

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Final answer:

The efficiency of an ideal gas engine is calculated using the net work done and the heat input during its thermodynamic cycle. The work for isothermal processes involves the natural logarithm of the volume change, and the Carnot efficiency, which cannot be determined without temperatures of the heat reservoirs, serves as a comparison point.

Step-by-step explanation:

The student is asking about calculating the efficiency of an engine that performs a cycle with an ideal gas going through four different thermodynamic processes. The engine cycle consists of isothermal expansion, cooling at constant volume, isothermal compression, and heating at constant volume. To find the efficiency, we need to consider both the work done and the heat exchanged during these processes.

To determine the efficiency, we use the formula:


Efficiency (ɳ) = (Work Output) / (Heat Input), where

  • Work Output is the net work done by the gas over the entire cycle.
  • Heat Input is the total heat absorbed during the isothermal expansion.

For an isothermal process, the work done by an ideal gas can be calculated using:


W = nRT ln(Vf/Vi), where

  • n is the number of moles.
  • R is the universal gas constant.
  • T is the temperature (constant during isothermal processes).
  • Vf is the final volume.
  • Vi is the initial volume.

Since the cycle returns to its original state, the change in internal energy (∆U) over the entire cycle is zero and thus the net work done is equal to the net heat exchanged (Qnet).

The efficiency of the engine can be compared to the Carnot efficiency which is calculated using temperatures of the two baths. However, this example does not provide specific temperatures for the heat reservoirs so a direct Carnot efficiency cannot be calculated from the given information.

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