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Gre4nikov [31]
1 year ago
5

Consider N non-interacting diatomic molecules stuck on a metal surface. Each molecule can either lie flat on the surface, in whi

ch case it must be aligned to one of two directions, x and y, or it can stand up along the z direction. There is an energy cost e > 0 associated with a molecule standing up, and zero energy for molecules lying flat along x or y directions.
What are the possible energy levels for this system of N particles, and what is the number of microstates associated to each energy level?
Find the entropy and energy of the system as a function of the temperature, T.
What is the probability of a specific molecule standing up?
What is the largest possible value of the energy of the system at any temperature?

Physics
1 answer:
Art [367]1 year ago
7 0

Answer:

Detailed solution given in diagram

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A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co
vaieri [72.5K]

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

6 0
2 years ago
What is the force on a .8 kg peach falling freely in a Yakima orchard
bezimeni [28]

Answer: Hence, ( 30,20 ) will not maximize the profit as it lies inside the solution region.

Explanation:

3 0
1 year ago
Read 2 more answers
An engineer wants to design an oval racetrack such that 3.20 × 10 3 lb racecars can round the exactly 1000 ft radius turns at 10
Reptile [31]

Answer:

The banking angle necessary for the race cars is 34.84°

Explanation:

For normal reaction the expression is:

\\Nsin\theta = \frac{mv^{2} }{R}  =Fc\\tan\theta =\frac{v^{2} }{Rg}  \\\theta =tan^{-1} (\frac{v^{2} }{Rg} )\\\theta =tan^{-1} (\frac{(102*0.447)^{2} }{1000*0.3048*9.8} )=34.84

4 0
1 year ago
The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end
Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

f_1=150\ Hz

for n=2

f_2=\frac{5v}{4L}

Thus there is one additional resonance corresponding to n=1 , between f_1 and f_2

8 0
2 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
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