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nikdorinn [45]
3 years ago
8

A 0.0434-m3 container is initially evacuated. Then, 4.19 g of water is placed in the container, and, after some time, all of the

water evaporates. If the temperature of the water vapor is 417 K, what is its pressure?Number units in Pa
Physics
1 answer:
adell [148]3 years ago
6 0

Answer:

P =18760.5 Pa

Explanation:

Given that

Volume ,V= 0.0434 m³

Mass ,m= 4.19 g = 0.00419 kg

T= 417 K

If we assume that water vapor is behaving like a ideal gas ,then we can use ideal gas equation

Ideal gas equation    P V = m R T

p=Pressure ,V = Volume ,m =mass

T=Temperature ,R=Universal gas constant

Now by putting the values

P V = m R T

For water R= 0.466 KJ/kgK

P x 0.0434 = 0.00419 x 0.466 x 417

P =18.7605 KPa

P =18760.5 Pa

Therefore the answer is 18760.5 Pa

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Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

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Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

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Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

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=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

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The minimun wall thickness:

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