The numerical value of the N v ( v ) N v ( v mp ) for the value of v = v m p 50.0 .

Question

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Answer 1

Question

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

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Answer 2

Question

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

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Answer 3

Question

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

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Answer 4

Question

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

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Answer 5

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

Answer 6

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Answer 7

Chapter 20, Problem 39AP

(a)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ .

Answer 8

(a)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Value of average speed is $vmp50.0$ .

Write the expression for the Maxwell-Boltzmann speed distribution function,

$Nv=4πN(m02πkBT)32v2e(−m0v22kBT)$ (1)

Here, $Nv$ is the Maxwell-Boltzmann speed distribution function, $N$ is the total number of molecules of gas, $T$ is the absolute temperature of gas, $v$ is the speed of the fraction of molecules of gas, $kB$ is the Boltzmann constant and $m0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=8kBTπm0$

Here, $v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$vmp=2kBTm0$

Here, $vmp$ is the most probable speed of a gas molecule.

Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$ using equation(1).

$Nv(v)Nv(vmp)=4πN(m02πkBT)32v2e(−m0v22kBT)4πN(m02πkBT)32vmp2e(−m0vmp22kBT)=(vvmp)2e(m0vmp22kBT−m0v22kBT)=(vvmp)2em0vmp22kBT(1−(vvmp)2)$ (2)

Substitute $2kBTm0$ for $vmp$ in equation (2) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vvmp)2em0(2kBTm0)22kBT(1−(vvmp)2)=(vvmp)2e(1−(vvmp)2)$ (3)

Conclusion:

Substitute $vmp50.0$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ ,

$Nv(v)Nv(vmp)=(vmp50.0vmp)2e(1−(vmp50.0vmp)2)=1.0868×10−3≈1.09×10−3$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp50.0$ is $1.09×10−3$ .

Answer 13

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Answer 14

(b)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ .

Answer 15

(b)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Value of average speed is $vmp10.0$ .

From equation (3), Write the formula to calculate the numerical value of the $Nv(v)Nv(vmp)$

$Nv(v)Nv(vmp)=(vvmp)2e(1−(vvmp)2)$

Conclusion:

Substitute $vmp10.0$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp10.0vmp)2e(1−(vmp10.0vmp)2)=2.69×10−2$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp10.0$ is $2.69×10−2$ .

Answer 20

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Answer 21

(c)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ .

Answer 22

(c)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Value of average speed is $vmp2.00$ .

Recall equation (3)

Conclusion:

Substitute $vmp2.00$ for $v$ in equation (3) to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmp2.00vmp)2e(1−(vmp2.00vmp)2)=0.529$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp2.00$ is $0.529$ .

Answer 27

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Answer 28

(d)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ .

Answer 29

(d)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Value of average speed is $vmp$ .

Recall equation (3)

Conclusion:

Substitute $vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(vmpvmp)2e(1−(vmpvmp)2)=1.00$

Therefore, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=vmp$ is $1.00$ .

Answer 34

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Answer 35

(e)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ .

Answer 36

(e)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Value of average speed is $2.00vmp$ .

Recall equation (3)

Conclusion:

Substitute $2.00vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(2.00vmpvmp)2e(1−(2.00vmpvmp)2)=0.199$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=2.00vmp$ is $0.199$ .

Answer 41

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Answer 42

(f)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ .

Answer 43

(f)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Value of average speed is $10.0vmp$ .

Recall equation (3),

Conclusion:

Substitute $10.0vmp$ for $v$ in above expression to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(10.0vmpvmp)2e(1−(10.0vmpvmp)2)=1.01×10−41$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=10.0vmp$ is $1.01×10−41$ .

Answer 48

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

Answer 49

(g)

To determine

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ .

Answer 50

(g)

Expert Solution

Answer to Problem 39AP

The numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Reacall equation (3)

Conclusion:

Substitute $50.0vmp$ for $v$ in above equation to find $Nv(v)Nv(vmp)$ .

$Nv(v)Nv(vmp)=(50.0vmpvmp)2e(1−(50.0vmpvmp)2)=1.25×10−1082$

Thus, the numerical value of the $Nv(v)Nv(vmp)$ for the value of $v=50.0vmp$ is $1.25×10−1082$ .

Answer 55

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The numerical value of the N v ( v ) N v ( v mp ) for the value of v = v m p 50.0 .

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