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2 years ago

What is the effect of the author proposing a "thought experiment" at the beginning of the text? FINANCIAL LITERACY

Physics

1 answer:

mojhsa [17]2 years ago

4 0

A thought experiment aims to promote speculative thinking, logical reasoning, and paradigm shifts. Thought experiments force us to address problems we find difficult to answer, which forces us out of our comfort zone. They highlight knowledge gaps and aid in recognizing the boundaries of what is knowable.

<h3>What are thought experiments?</h3>

Thought experiments are imaginative methods for examining the nature of things. They are employed for a variety of purposes in a number of disciplines, including physics, economics, history, mathematics, and philosophy.

A thought experiment is a test that we do in our minds. Here, we will imagine a specific scenario, consider some of its repercussions, and come to a broad conclusion.

Therefore, a thought experiment aims to promote speculative thinking, logical reasoning, and paradigm shifts. Thought experiments force us to address problems we find difficult to answer, which forces us out of our comfort zone. They highlight knowledge gaps and aid in recognizing the boundaries of what is knowable.

Learn more about experiments on:

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he magnetic flux through a loop of wire decreases from 1.7 Wb to 0.3 Wb in a time of 0.4 s. What was the average value of the in

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Explanation:

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Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at –108.0 °C. What is the vapor pressure o

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Answer:

P₁ = 0.0562 atm

Explanation:

Using the Clausius-Clapeyon equation

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁) ------ (eqn 1)

Step 1: From the question given, we state out the parameters given

P₁ = ? T₁ = -148.0⁰C

P₂ = 1atm T₂ = -108.0⁰C

ΔHvap = 12.6kJ/mol R = 8.314J/K.mol

Step 2: Do conversions where necessary for unit consistency since our R value is in J/K.mol

a) convert ⁰C to K

1K = ⁰C + 273.15

T₁ = -148.0⁰C => -148.0⁰C + 273.15

T₁ = 125.15K

T₂ = -108.0⁰C => -108.0⁰C + 273.15

T₂ = 166.15K

b) convert kJ/mol to Joules

ΔHvap = 12.6kJ/mol = 12600Joules

substituting parameters into eqn 1

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁)

ln (P₁/1atm) = 12600J / 8.314 (1/166.15 - 1/125.15)

= 1515.51 (0.0060 - 0.0079)

= 1515.51(-0.0019)

ln (P₁/1atm) = -2.8794

taking exponential of both sides to get rid of the natural log

P₁ = e^ -2.8794

P₁ = 0.05616 atm

P₁ = 0.0562atm

Key Words

1) Clausius-Clapeyon: shows the relationship between pressure and temperature and it is used to estimate the vapour of a solution at a different temperature

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