Home
Class 12
CHEMISTRY
Calculate rms speed of O(2) at 273 K and...

Calculate rms speed of `O_(2)` at 273 K and `1 xx 10^(5)` Pa pressure. The density of `O_(2)` under these conditions is `1.42 kg m^(-3)`.

Text Solution

Verified by Experts

Data are given in SI units
`C= sqrt( ( 3P)/( d)) = sqrt((3 xx 10^(5))/( 1.42)) = 459.63 m sec^(-1)`
Promotional Banner

Topper's Solved these Questions

  • GASEOUS STATE

    MOTION|Exercise SOLVED EXAMPLE|15 Videos

  • GASEOUS STATE

    MOTION|Exercise SOLVED EXAMPLE(IIT-JEE MAINS)|40 Videos

  • ELECTROCHEMISTRY

    MOTION|Exercise EXERCISE-4,II|44 Videos

  • GOC

    MOTION|Exercise Exercise - 4 Level - II|14 Videos

Similar Questions

Explore conceptually related problems

Calculate the rms speed of nitrogen at STP (pressure =1 atm and temperature = 0^0 C . The density of nitrgoen in these conditions is 1.25 kg m^(-3)

N_(2)O_(4) is 60% dissociated into NO_(2) at 340K and 1 atm pressure. Find the volume of 10 g N_(2)O_(4) occupy under these condition.

The osmotic pressure of a solution at 273 K is 2.5 atm. Its osmotic pressure at 546 K under similar conditions will be :

At a temperature under high pressure K_(w)(H_(2)O)=1 xx 10^(-10) A solution of pH 5.4 under these conditions is sait to be :

Calculate the rms speed of smoke particles of mass 5 xx 10^(-17) kg in their Brownian motion in air at NTP. Given k_(B) = 1.38 xx 10^(-23) J//K

At 340 K and 1 atm pressure, N_(2)O_(4) is 66% into NO_(2) . What volume of 10 g N_(2)O_(4) ocuupy under these conditions?

At 273.15K and under a pressure of 10.1325 Mpa, the compressibility factor of O_(2) is 0.927. Calcualte the mass of O_(2) necessary to fill a gas cylinder of 100 dm^(3) capacity under the given conditions.

Calculate the depth of water at which a bubble of air would just float. Density of water =1000 kg^(-3) and density of air = 1.29 kg ^(-3) . Atmospheric pressure = 10^(5) pa.

MOTION-GASEOUS STATE -EXERCISE-4( LEVEL-II)
  1. Calculate rms speed of O(2) at 273 K and 1 xx 10^(5) Pa pressure. The ...

    Text Solution

    |

  2. The average velocity of gas molecules is 400 m/sec calculate its rms v...

    Text Solution

    |

  3. For one mole of gas the average kinetic energy is given as E.The U("rm...

    Text Solution

    |

  4. Ratio of rates of diffusion of He and CH(4) (under identical condition...

    Text Solution

    |

  5. Figure displays the plot of the compression factor Z verses pfor a few...

    Text Solution

    |

  6. Match gases under specified condition listed in Column I with their pr...

    Text Solution

    |

  7. A gas described by van der Waals equation

    Text Solution

    |

  8. The term that corrects for the attractive forces present in a real gas...

    Text Solution

    |

  9. At 400 K, the root mean square (rms) speed of a gas X (molecular weigh...

    Text Solution

    |

  10. According to kinetic theory of gases:

    Text Solution

    |

  11. To an evacuated vessel with movable piston under external pressure of ...

    Text Solution

    |

  12. For one mole of a van der Waals gas when b =0 and T =30 K the PV vs1//...

    Text Solution

    |

  13. The atomic masses of He and Ne are 4 and 20 amu respectively . The va...

    Text Solution

    |

  14. A fixed mass m of a gas is subjected to transformation of state: K to ...

    Text Solution

    |

  15. A fixed mass 'm' of a gas is subjected to transformation of states fro...

    Text Solution

    |

  16. X and Y are two volatile liquids with molar weights of 10gmol^(-1) and...

    Text Solution

    |

  17. X and Y are two volatile liquids with molar weights of 10gmol^(-1) and...

    Text Solution

    |

  18. One mole of a monoatomic real gas satisfies the equation p(V-b)=RT wh...

    Text Solution

    |

  19. The diffusion coefficient of an ideal gas is proportional to its mean ...

    Text Solution

    |

  20. A closed tank has two compartments A and B, both filled with oxygen (a...

    Text Solution

    |