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Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found to be 1150 m/s. What temperature does this represent

Respuesta :

Answer:

[tex]T=954.41\ K[/tex]

Explanation:

The expression for the root mean square speed is:

[tex]C_{rms}=\sqrt {\dfrac {3RT}{M}}[/tex]

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

Molar mass of water vapor = 18.0 g/mol  = 0.018 kg/mol

Temperature = ?

[tex]C_{rms}=1150[/tex]  m/s

[tex]1150=\sqrt{\frac{3\times 8.314\times T}{0.018}}[/tex]

[tex]1322500=\frac{24.942T}{0.018}[/tex]

[tex]T=954.41\ K[/tex]

Cricetus

This represent by the temperature "954.41 K".

Given values:

  • Molar mass = 18.0 g/mol or, 0.018 kg/mol
  • Root mean square velocity, [tex]C_{rms} = 1150 \ m/s[/tex]
  • Gas constant, [tex]R = 8.314 \ J/K.mol[/tex]

As we know the relation,

→     [tex]C_{rms} = \sqrt{\frac{3RT}{M} }[/tex]

By substituting the values, we get

→      [tex]1150 = \sqrt{\frac{3\times 8.314\times T}{0.018} }[/tex]

→ [tex]1322500 = \frac{24.942 \ T}{0.018}[/tex]                        

→          [tex]T = 954.41 \ K[/tex]

Thus the above approach is right.  

Learn more about temperature here:

https://brainly.com/question/15995507

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