3. 0.050 moles of a monatomic gas expands adiabatically and quasistatically from 1.00 liters to 2.00 liters. The initial pressure of the gas is 155 kPa. (a) What is the initial temperature of the gas?

Answers

Answer 1

The initial temperature of the gas is 374 K or 101°C approximately.

Given that the amount of a monatomic gas is 0.050 moles which is expanding adiabatically and quasistatically from 1.00 L to 2.00 L.

The initial pressure of the gas is 155 kPa. We have to calculate the initial temperature of the gas. We can use the following formula:

PVγ = Constant

Here, γ is the adiabatic index, which is 5/3 for a monatomic gas. The initial pressure, volume, and number of moles of gas are given. Let’s use the ideal gas law equation PV = nRT and solve for T:

PV = nRT

T = PV/nR

Substitute the given values and obtain:

T = (155000 Pa) × (1.00 L) / [(0.050 mol) × (8.31 J/molK)] = 374 K

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Related Questions

3. Consider a 7-DOF system with mass matrix [M] and stiffness matrix [K]. A friend has discovered three vectors V₁, V₂ and V3 such that VT[M]V₁ = 0 VT[K]V₁ = 0 forij. Has your friend found 3 eigenvectors of the system? Do you need any more information? What else can you tell your friend about these vectors?

Answers

Yes, your friend has found 3 eigenvectors of the system. An eigenvector is a vector that, when multiplied by a matrix, produces a scalar multiple of itself.

In this case, the vectors V₁, V₂, and V₃ are eigenvectors of the system because, when multiplied by the mass matrix [M] or the stiffness matrix [K], they produce a scalar multiple of themselves.

I do not need any more information to confirm that your friend has found 3 eigenvectors. However, I can tell your friend a few things about these vectors. First, they are all orthogonal to each other. This means that, when multiplied together, they produce a vector of all zeros. Second, they are all of unit length. This means that their magnitude is equal to 1.

These properties are important because they allow us to use eigenvectors to simplify the analysis of a system. For example, we can use eigenvectors to diagonalize a matrix, which makes it much easier to solve for the eigenvalues of the system.

Here are some additional details about eigenvectors and eigenvalues:

An eigenvector of a matrix is a vector that, when multiplied by the matrix, produces a scalar multiple of itself.

The eigenvalue of a matrix is a scalar that, when multiplied by an eigenvector of the matrix, produces the original vector.

The eigenvectors of a matrix are orthogonal to each other.

The eigenvectors of a matrix are all of unit length.

Eigenvectors and eigenvalues can be used to simplify the analysis of a system.

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Show that the free-particle one-dimensional Schro¨dinger
equation for the wavefunc-
tion Ψ(x, t):
∂Ψ
i~
∂t = −
~
2
2m


,
∂x2
is invariant under Galilean transformations
x
′ = x −
3. Galilean invariance of the free Schrodinger equation. (15 points) Show that the free-particle one-dimensional Schrödinger equation for the wavefunc- tion V (x, t): at h2 32 V ih- at is invariant u

Answers

The Galilean transformations are a set of equations that describe the relationship between the space-time coordinates of two reference systems that move uniformly relative to one another with a constant velocity. The aim of this question is to demonstrate that the free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is invariant under Galilean transformations.

The free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is represented as:$$\frac{\partial \psi}{\partial t} = \frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$Galilean transformation can be represented as:$$x' = x-vt$$where x is the position, t is the time, x' is the new position after the transformation, and v is the velocity of the reference system.

Applying the Galilean transformation in the Schrodinger equation we have:

[tex]$$\frac{\partial \psi}{\partial t}[/tex]

=[tex]\frac{\partial x}{\partial t} \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial t}$$$$[/tex]

=[tex]\frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$[/tex]

Substituting $x'

= [tex]x-vt$ in the equation we get:$$\frac{\partial \psi}{\partial t}[/tex]

= [tex]\frac{\partial}{\partial t} \psi(x-vt, t)$$$$\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x} \psi(x-vt, t)$$$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]

Substituting the above equations in the Schrodinger equation, we have:

[tex]$$\frac{\partial}{\partial t} \psi(x-vt, t) = \frac{-\hbar}{2m} \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]

This shows that the free-particle one-dimensional Schrodinger equation is invariant under Galilean transformations. Therefore, we can conclude that the Schrodinger equation obeys the laws of Galilean invariance.

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The end of the cylinder with outer diameter = 100 mm and inner diameter =30 mm and length = 150 mm will be machined using a CNC lathe machine with rotational speed =336 rotations per minute, feed rate = 0.25 mm/ rotation, and cutting depth = 2.0 mm. Machine mechanical efficiency =0.85 and specific energy for Aluminum = 0.7 N−m/m³. Determine: i. Cutting time to complete face cutting operation (sec). ii. Material Removal Rate (mm³/s). iii. Gross power used in the cutting process (Watts).

Answers

i. Cutting time: Approximately 53.57 seconds.

ii. Material Removal Rate: Approximately 880.65 mm³/s.

iii. Gross power used in the cutting process: Approximately 610.37 Watts.

To determine the cutting time, material removal rate, and gross power used in the cutting process, we need to calculate the following:

i. Cutting time (T):

The cutting time can be calculated by dividing the length of the cut (150 mm) by the feed rate (0.25 mm/rotation) and multiplying it by the number of rotations required to complete the operation. Given that the rotational speed is 336 rotations per minute, we can calculate the cutting time as follows:

T = (Length / Feed Rate) * (1 / Rotational Speed) * 60

T = (150 mm / 0.25 mm/rotation) * (1 / 336 rotations/minute) * 60

T ≈ 53.57 seconds

ii. Material Removal Rate (MRR):

The material removal rate is the volume of material removed per unit time. It can be calculated by multiplying the feed rate by the cutting depth and the cross-sectional area of the cut. The cross-sectional area of the cut can be calculated by subtracting the area of the inner circle from the area of the outer circle. Therefore, the material removal rate can be calculated as follows:

MRR = Feed Rate * Cutting Depth * (π/4) * (Outer Diameter^2 - Inner Diameter^2)

MRR = 0.25 mm/rotation * 2.0 mm * (π/4) * ((100 mm)^2 - (30 mm)^2)

MRR ≈ 880.65 mm³/s

iii. Gross Power (P):

The gross power used in the cutting process can be calculated by multiplying the material removal rate by the specific energy for aluminum and dividing it by the machine mechanical efficiency. Therefore, the gross power can be calculated as follows:

P = (MRR * Specific Energy) / Machine Efficiency

P = (880.65 mm³/s * 0.7 N−m/m³) / 0.85

P ≈ 610.37 Watts

So, the results are:

i. Cutting time: Approximately 53.57 seconds.

ii. Material Removal Rate: Approximately 880.65 mm³/s.

iii. Gross power used in the cutting process: Approximately 610.37 Watts.

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