The mean speed of the emerging beam relative to the initial value has to be calculated. Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as, v mean = ( 8 R T π M ) 1 / 2

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Answer 1

Question

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

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Answer 2

Question

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

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Answer 3

Question

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

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Answer 4

Question

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

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Answer 5

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

Answer 6

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).

$vx=(2kTmπ)1/2×1−e−1/πerf(1/π1/2)$ (5)

The value of $erf(1/π1/2)$ is,

$erf(1/π1/2)=erf(0.56)=0.57$

The value of $e−1/π$ is $0.73$ .

Substitute the values in equation (5) to calculate mean velocity.

$vx=(2kTmπ)1/2(1−0.730.57)=0.47〈vx〉initial_$

Thus, the value of mean velocity of emerging beam is $vx=0.47〈vx〉initial_$ .

Answer 7

Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution. The mean speed for the gas particles is represented as,

$vmean= (8RTπM)1/2$

Answer 8

Expert Solution & Answer

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $vx=0.47〈vx〉initial_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The expression for the velocity of the emerging beam is,

$vx=K∫0avxf(vx)dvx$ (1)

Where,

• $K$ is the constant of proportionality.
• $vx$ is the velocity.

The value of $f(vx)$ is given by,

$f(vx)=(m2πkT)1/2e−mv2/2kT$

Where,

• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The normalised function is written as,

$vx∫0avxdvx=K∫0af(vx) dvx1=K∫0af(vx) dvx$

Substitute the value of $f(vx)$ in the above equation.

$1=K(m2πkT)1/2∫0ae−mv2/2kT dvx$ (2)

The equation (2) is evaluated by defining an error function.

$mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n$

Or,

$dvx=(2kTm)1/2dn$ (3)

Substitute equation (3) in equation (2) to find $n$ .

$1=K(m2πkT)1/2(2kTm)1/2∫0be−n2dn$

Here, $b=(m2kT)1/2×a$

$1=K(2kTm)1/2(m2πkT)1/2∫0be−n2dn=(Kπ)1/2∫0be−n2dn$

According to error function,

$erf(z)=2π1/2∫0ze−n2dn$

Therefore,

$1=(Kπ)1/2∫0be−n2dn=12K erf(b)=2erf(b)$

To calculate the mean velocity of the beam, substitute the values in equation (1).

$vx=K(m2πkT)1/2∫0avxe−mvx2/2kTdvx=K(m2πkT)1/2(−kTm)∫0addvx(e−mvx2/2kTdvx)$

$vx=−(kT2mπ)1/2(e−mvx2/2kT−1)$ (4)

The initial velocity, $a=〈vx〉initial$ and the initial velocity is $(2kT/mπ)1/2$ . The expression for average magnitude of one dimensional velocity in the $x$ direction is,

$〈vx〉initial=2∫0∞vxf(vx) dvx=2∫0∞vx(m2πkT)1/2 e−mv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2$

For $a=∞$ the value of constant, $b=∞$ and the value of error function, $b erf(b)=1$ . Substitute the value of $a$ and $b erf(b)=erf(1/π1/2)$ in equation (4).