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Chapter 19: Problem 40

A reversible engine contains 0.20 mol of ideal monatomic gas, initially at \(600 \mathrm{K}\) and confined to \(2.0 \mathrm{L} .\) The gas undergoes the following cycle: Isothermal expansion to \(4.0 \mathrm{L}\) \(\cdot\) Isovolumic cooling to \(300 \mathrm{K}\) Isothermal compression to \(2.0 \mathrm{L}\) \(\cdot\) Isovolumic heating to \(600 \mathrm{K}\) (a) Calculate the net heat added during the cycle and the net work done. (b) Determine the engine's efficiency, defined as the ratio of the work done to the heat absorbed during the cycle.

Short Answer

Expert verified
The net work done during the cycle is 276.35 J, the net heat added is 276.35 J, and the efficiency of the engine is 1 or 100%.

Step by step solution

01

Calculate the work done during the isothermal expansion

During an isothermal process (constant temperature), the work done by the gas can be calculated using the formula \(W = nRT \log(\frac{V_{f}}{V_{i}})\), where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature, and \(V_{i}\) and \(V_{f}\) are the initial and final volumes, respectively. For the isothermal expansion from 2.0 L to 4.0 L at 600 K, this gives \(W_{1} = 0.20 mol \times 8.314 J/(mol K) \times 600 K \times \log(\frac{4.0 L}{2.0 L}) = 552.69 J\). Note that the work done by the system is positive as it is expanding.
02

Calculate the work done during the isovolumic process

During an isovolumic process (constant volume), there is no work done by the gas as the volume doesn’t change. Thus \(W_{2} = 0\).
03

Calculate the work done during the isothermal compression

It is the same process as in step 1, but in this case, the gas is being compressed from 4.0 L to 2.0 L at a temperature of 300 K. Work done \(W_{3} = 0.20 mol \times 8.314 J/(mol K) \times 300 K \times \log(\frac{2.0 L}{4.0 L}) = -276.34 J\). It's negative because work is done on the system.
04

Calculate the work done during the second isovolumic process

Just like in step 2, \(W_{4} = 0\).
05

Calculate the total work done

The total work done is the sum of the work done in each step: \(W = W_{1} + W_{2} + W_{3} + W_{4} = 552.69 J - 276.34 J = 276.35 J\).
06

Calculate the net heat added during the cycle

The change in internal energy \(\Delta U\) for one complete cycle is 0 since the gas returns to its initial state. From the first law of thermodynamics \(\Delta U = Q - W\), where \(Q\) is the heat added to the system and \(W\) is the work done by the system. Therefore, the net heat added during the cycle is \(Q = \Delta U + W = 0 + 276.35 J = 276.35 J\).
07

Determine the engine's efficiency

The efficiency of the engine is the ratio of the work done to the heat absorbed during the cycle. Thus, the efficiency is \(\frac{W}{Q} = \frac{276.35 J}{276.35 J} = 1\). However, in a real engine process, such 100% efficiency is not possible due to non-ideal conditions.

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