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nalin [4]
2 years ago
12

A constant volume perfect gas thermometer indicates a pressure of 6.69 kPaat the triple point of water (273.16 K). (a) What chan

ge of pressure indicates a change of 1.0 K at this temperature? (b) What pressure indicates a temperature of 100 ⁰C? (c) What change of pressure indicate a change of 1.0 K at the latter temperature?
Physics
1 answer:
madam [21]2 years ago
7 0

Answer:

(a) = 6.714 kPa

(b) = 9.14 kPa

(c) = 2.47 kPa

Explanation:

Pressure Law: Pressure law states that the pressure of a fixed mass of gas is directly proportional to its absolute temperature provided the volume is kept constant. It is expressed mathematically as

P₁/T₁ = P₂/T₂

Triple point of water: The triple point of water is the temperature at which water is in equilibrium with the three phases ( solid, liquid and gaseous state.)

Applying Pressure law,

P₁/T₁ = P₂/T₂ .......................... Equation 2

Where P₁ = Initial pressure, T₁ = initial Temperature, P₂ = final Temperature, T₂ = Final Temperature.

(a) At 1.0 K change in temperatures

Making P₂ The subject of the equation in Equation 2,

P₁ = 6.69 kPa = 6.69 × 10³ Pa, T₁ = 273.16 K, T₂ = 273.16 + 1 (change in temperature of 1 .0 K) = 274.16 K.

Substituting these values into equation 2

P₂ =( P₁ × T₂)/T₁

P₂ = (6.69 × 274.16)/273.16

P₂ = 1834.13/273.16

P₂  =6.714 kPa

(b) At a temperature of 100 ⁰C.

P₂ = ( P₁ × T₂)/T₁

Where P₁ = 6.69 Kpa, T₁ = 273.16 K, T₂ = 100 + 273 = 373 K

Substituting these values into the equation above,

∴ P₂ = (6.69×373)/273.16

  P₂ = 2495.37/273.16

   P₂ = 9.14 kPa

(c)P₂ = ( P₁ × T₂)/T₁

Where,  P₁ = 6.69 Kpa, T₁ = 273.16K, T₂ =  373 + 1 = 374 K

∴ P₂ = (6.69 × 374)/273.16

   P₂ = 2502.06/273.16

   P₂ = 9.16 kPa.

∴change in pressure (ΔP) = P₂ - P₁

    (ΔP)  = 6.69 - 9.16 = 2.47 kPa.

   

 

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All of the above mentioned ranges are compared to that of humans.

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Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
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The speed of a proton after it accelerates from rest through a potential difference of 350 V is 25.86 \times 10^4 ~m/s.

Initial velocity of the proton u = 0

Given potential difference \Delta V = 350V

let's assume that the speed of the proton is v,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge q when accelerated with a potential difference \Delta V is,

    U = q \Delta V

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy  P.E must be equal to gain in Kinetic Energy K.E</em> i.e

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If the initial and final velocity of the proton is u and v respectively then,

change in Kinetic Energy  \implies  \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0

change in Potential Energy \implies \Delta U = q\Delta V

from conservation of energy,

             v= \sqrt{\frac{2q\Delta V}{m}}

so,         v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}

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7 0
1 year ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
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Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
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Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

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The Electric field intensity due to a charge is given as:

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords:  Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0
3 years ago
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