Home
Class 11
CHEMISTRY
Calculate the average kinetic energy in ...

Calculate the average kinetic energy in joules of the molecules in 8.0 g of methane at `27^@C`.

Text Solution

Verified by Experts

Total kinetic energy `=n((3)/(2)RT)`
when `n` is the number of moles
Moles of methane `=(W)/(Molar weight)=(8)/(16)=0.5 mol`
`:. KE=0.5xx(3)/(2)xx8.314xx300=1870.65 J`
Therefore, average kinetic energy is
`=(1870.65)/(6.023xx10^(23)xx0.5)=6.21xx10^(-21)J`
Promotional Banner

Topper's Solved these Questions

  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Illustration|2 Videos
  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises|21 Videos
  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|11 Videos
  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|33 Videos

Similar Questions

Explore conceptually related problems

The total kinetic energy in joules of the molecules in 8 g of methane at 27^@C is

The average kinetic energy in Joule of the molecules in 8 grams of CH_(4) at 27^(@)C is

Calculate the average kinetic energy of a hydrogen molecule at 27^@C .

Given, Avogadro's number N = 6.02 xx 10^23 and Boltzmann's constant k = 1.38 xx 10^-23 J//K . (a) Calculate the average kinetic energy of translation of the molecules of an ideal gas at 0^@ C and at 100^@ C . (b) Also calculate the corresponding energies per mole of the gas.

The average kinetic energy of a gas molecule is

The number of molecules in 16g of methane is:

Calculate the average kinetic energy of one mole of CO_(2) at 450 K in Joules.

What happens to the average kinetic energy of the molecules as ice melts at 0^(@)C ?

The kinetic energy of 1 mole of oxygen molecules in cal mol^(-1) at 27°C

Calculate the total and average kinetic energy of 32 g methane molecules at 27^(@)C (R =8.314 JK^(-1) mol^(-1)) .

CENGAGE CHEMISTRY ENGLISH-STATES OF MATTER-Exercises (Ture False)
  1. Calculate the average kinetic energy in joules of the molecules in 8.0...

    Text Solution

    |

  2. In the van der Waals equation (P + (n^(2)a)/(V^(2)))(V - nb) = nRT ...

    Text Solution

    |

  3. Kinetic energy of a molecule is zero at 0^(@)C

    Text Solution

    |

  4. Gas in a closed container will exert much higher pressure due to gravi...

    Text Solution

    |

  5. The graph between PV vs P at constant temperature is linear parallel t...

    Text Solution

    |

  6. Real gases show deviation from ideal behavior at low temperature and h...

    Text Solution

    |

  7. All the molecules in a given sample of gas move with same speed.

    Text Solution

    |

  8. Small value of a means, gas can be easily liqueifed.

    Text Solution

    |

  9. Small value of a means, gas can be easily liqueifed.

    Text Solution

    |

  10. Rate of diffusion is directly proportional to the square root of molec...

    Text Solution

    |

  11. For ideal gases, Z = 1 at all temperature and pressure.

    Text Solution

    |

  12. According to charles's law,

    Text Solution

    |

  13. The pressure of moist gas is higher than pressure of dry gas.

    Text Solution

    |

  14. Gases do not occupy volume and do not have force of attraction.

    Text Solution

    |

  15. The van der Waal equation of gas is (P + (n^(2)a)/(V^(2))) (V - nb)...

    Text Solution

    |

  16. Surface tension and surface energy have different dimensions.

    Text Solution

    |

  17. The plot of PV vs P at particular temperature is called isovbar.

    Text Solution

    |

  18. Give reasons for the following in one or two sentences. (a) A bottle...

    Text Solution

    |

  19. Can a gas with a = 0 be liquefied?

    Text Solution

    |

  20. The van der waals constants have same values for all the gases.

    Text Solution

    |

  21. All the molecules in a given sample of gas move with same speed.

    Text Solution

    |