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nalin [4]
3 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]3 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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According to the physics constants table on Chemistry Libretexts:

  • Proton rest mass: \rm 1.0072765\;amu;
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<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other 56 - 26 = 30 particles are neutrons.

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\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu.

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

\rm 56.464736 - 55.934939 = 0.514197\;amu.

<h3>b)</h3>

By the mass-energy equivalence,

E = m\cdot c^{2}.

Refer to this equation, the speed of light in vacuum c^{2} is the conversion factor between mass m and energy E. The value of c is usually given only in SI units \rm m\cdot s^{-1}. Accordingly, the value of c^{2} will be in the SI unit \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}.

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.  

\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}.

Convert the unit of c^{2} from \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1} to the desired \rm MeV \cdot amu^{-1}:

\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}.

Total binding energy in each iron-56 nucleus:

\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}.

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 \mathrm{^{56}Fe} will be:

\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV.

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