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Leokris [45]
2 years ago
7

Given the scope of Kayla’s education, training, and years of experience as a CMA, would this “favor” fall within the AAMA guidel

ines of her responsibilities?
Would any portion of Dr. Hsu’s request fall within the guidelines? If so, which portion(s)? Is an exception to these guidelines ever allowed?

How should Kayla respond to Dr. Hsu?
Physics
1 answer:
Cerrena [4.2K]2 years ago
6 0

Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.

From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.

Even though the favor that was asked by Dr. Hsu was prohibited by AAMA, it should be noted that the final part of the favor about faxing the report to the internist would fall within AAMA guidelines.

In conclusion, the best way that Kayla can respond to Dr. Hsu is to decline doing the favor.

Read related link on:

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Explanation:

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Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

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or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

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(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

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(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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