According to Boyle's Law, if the temperature of a confined gas is held fixed, then the product of the pressure P and the volume V is a constant, suppose that, for a certain gas, PV=800 where P is measured in pounds per square inch and V is measured in cubic inches. Answer the following: (this is calculus)
A) Find the average rate of change of P as V increases from 200in^3 to 250in^3.
B) Express V as a function of P and show that the instantaneous rate of change of V with respect to P is inversely proportional to the square of P.

Answers

Answer 1

If PV = 800, then P can be written as a function of V,

P(V) = 800 / V

(a) The average rate of change of P as V increases from 200 to 250 in³ is then

(P (250) - P (200)) / (250 in³ - 200 in³) = (3.2 lb/in² - 4lb/in²) / (50 in³)

... = -0.016 (lb/in²)/in³

(Or -0.016 lb/in⁵, but I figure writing the rate as (units of pressure) per (unit volume) makes more sense.)

(b) We can also write V as a function of P :

V(P) = 800 / P

Take the derivative:

V'(P) = - 800 / P²

which immediately demonstrates that V'(P) ∝ 1 / P², as required. (The fish-looking symbol, ∝, means "is proportional to".)

If differentiating is supposed to be more involved, you can use the limit definition:

[tex]V'(P)=\displaystyle\lim_{h\to0}\frac{V(P+h)-V(P)}h[/tex]

[tex]V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800}{P+h}-\frac{800}P}h[/tex]

[tex]V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800P-800(P+h)}{P(P+h)}}h[/tex]

[tex]V'(P)=\displaystyle800\lim_{h\to0}\frac{-\frac h{P(P+h)}}h[/tex]

[tex]V'(P)=\displaystyle-800\lim_{h\to0}\frac1{P(P+h)}=-\dfrac{800}{P^2}[/tex]

Answer 2

Part(a): The average rate of change of [tex]\frac{dP}{dV}[/tex] is [tex]-0.0164 lb/in^{2}[/tex]

Part(b): The required answer is [tex]\frac{dV}{dP} \infty \frac{1}{P^{2}}[/tex]

Part(a):

Given,

[tex]PV=800[/tex]

Differentiating the above equation with respect to [tex]V[/tex].

[tex]\frac{dP}{dV}=-\frac{P}{V}[/tex]

Now, at [tex]V=200 in^{3}[/tex] and [tex]P=4 lb/in^2[/tex]

At [tex]V=250 in^3[/tex] and [tex]P=\frac{80}{25} lb/in^2[/tex]

Hence, we can write,

[tex]\frac{dP}{dV}[/tex] at [tex]V=200[/tex] we get,

[tex]\frac{-1}{200}=\frac{-1}{50}[/tex]

Again [tex]\frac{dP}{dV}[/tex] at [tex]V=250[/tex] we get,

[tex]\frac{-80}{25\times 250}[/tex]

So, the average rate of change of [tex]\frac{dP}{dV}[/tex] is,

[tex]\frac{-\frac{1}{50}- \frac{8}{25\times25} }2\\=-\frac{1}{100}-\frac{1}{625}\\=-0.0164 lb/in^2[/tex]

Part(b):

Given,[tex]PV=800[/tex] then,

[tex]V=\frac{800}{P}[/tex]

Now,[tex]\frac{dV}{dP} =-\frac{800}{P^2}[/tex]

Then the above implies,

[tex]\frac{dV}{dP} \infty \frac{1}{P^{2}}[/tex]

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"PEDMAS is an order of operation used in mathematics to deal easily with complex calculations. It states that we start solving any arithmetic expression by solving the terms written in parentheses or brackets and then we simplify exponential terms and move ahead to division and multiplication operations and then, at last, we can find the answer by solving addition and subtraction operations."

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Answers

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Step-by-step explanation:

16. A section of an exam contains four True-False questions. A completed exam paper is selected at random, and the four answers are recorded. a. List all outcomes in the sample space. b. Assuming the outcomes to be equally likely, find the probability that all the answers are the same. c. Assuming the outcomes to be equally likely, find the probability that exactly one of the four answers is "True." d. Assuming the outcomes to be equally likely, find the probability that at most one of the four answers is "True."

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Answer:

a

[ FFFF] , [FFFT] , [FFTF] , [FFTT]

[FTFF] , [FTFT] , [FTTF], [FTTT]

[TFFF] , [TFFT] , [TFTF] , [TFTT]

[TTFF] , [TTFT],[TTTF] , [TTTT]

b

  [tex]P(A) =  \frac{1}{8}[/tex]

c

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d

[tex]P(C) = \frac{5}{16}[/tex]

Step-by-step explanation:

From the question we are told that

The number of True - False question is n = 4

The number of answers recorded are k = 4

Generally the outcomes in the sample space are 16 and they are listed below

[ FFFF] , [FFFT] , [FFTF] , [FFTT]

[FTFF] , [FTFT] , [FTTF], [FTTT]

[TFFF] , [TFFT] , [TFTF] , [TFTT]

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=> [tex]P(A) = \frac{1}{8}[/tex]

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[tex]P(B) = \frac{4}{16}[/tex]

=> [tex]P(B) = \frac{1}{4}[/tex]

Generally from the list of the possible outcome we see that the number of outcome where at most one of the four answers is true is 5 i.e [FTFF] , [FFFF] , [FFTF] , [FFFT] ,[TFFF]

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[tex]P(C) = \frac{5}{16}[/tex]

The probabilities of all same, exactly one true, and most one of the four are true is 0.125, 0.25, and 0.3125 respectively.

What is probability?

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

A section of an exam contains four True-False questions.

A completed exam paper is selected at random, and the four answers are recorded.

Total event = 16

A.  The sample space will be shown in the table below.

[tex]\rm Sample \ space = 16 \ \begin{Bmatrix}FFFF & FFFT & FFTF & FFTT \\FTFF & FTFT & FTTF & FTTT \\TFFF & TFFT & TFTF & TFTT \\TTFF & TTFT & TTTF & TTTT\end{Bmatrix}[/tex]

B.  The probability that all the answers are the same will be.

Favorable event = 2 ={(TTTT, FFFF}

Then we have the probability,

[tex]\rm P(B) =\dfrac{2}{16} =\dfrac{1}{8} = 0.125[/tex]

C.  The probability that exactly one of the four answers is "True".

Favorable event = 4 {(FFFT, FFTF, FTFF, TFFF}

Then we have the probability,

[tex]\rm P(C) =\dfrac{4}{16} =\dfrac{1}{4} = 0.25[/tex]

D.  The probability that at most one of the four answers is "True".

Favorable event = 5 {(FFFF, FFFT, FFTF, FTFF, TFFF}

Then we have the probability,

[tex]\rm P(D) =\dfrac{5}{16} =0.3125[/tex]

More about the probability link is given below.

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