Home
Class 11
CHEMISTRY
The volume percentages of N(2),O(2) and ...

The volume percentages of `N_(2),O_(2)` and He in gas mixture are 25, 35 and 40 , respectively. At a given temperature, the pressure of the mixture is 760 mm Hg. Calculate the partial pressure of each gas at the same temperature.

Text Solution

Verified by Experts

According to Amagat's law, at constant temperature and pressure, the partial volume of a component gas in a gas mixture=molee-fractino of the compound`xx`total volume of the mixture.
therefore, in the given mixture`,x_(N_(2))=(25)/(100)=0.25`,
`x_(O_(2))=(35)/(100)=0.35 and x_(He)=(40)/(100)=0.40`
`therefore `In mixture,
Partial pressure of `N_(2)=x_(N_(2))xxP=0.25xx760=190` mm Hg
Partial pressure of `O_(2)=x_(X_(2))xxP=0.35xx760`=266 mm Hg
and partial pressure of `He=x_(He)xxP=0.40xx760`=304 mm Hg
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • STATES OF MATTER : GASES AND LIQUIDS

    CHHAYA PUBLICATION|Exercise WARM UP EXERCISE|104 Videos

  • STATES OF MATTER : GASES AND LIQUIDS

    CHHAYA PUBLICATION|Exercise QUESTION-ANSWER ZONE FOR BOARD EXAMINATION|51 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    CHHAYA PUBLICATION|Exercise PRACTICE SET|13 Videos

  • STRUCTURE OF ATOM

    CHHAYA PUBLICATION|Exercise PRACTICE SET|15 Videos

Similar Questions

Explore conceptually related problems

Volumes of N_(2) and O_(2) in any gas mixture are 80% and 20T respectively. Determine the average vapour density of the gas mixture.

A mixture of O_(2) and H_(2) gases contains 20% of H_(2) gas. At a certain temperature, the total pressure of the mixture is found to be 1 bar. What is the partial pressure of O_(2) (in bar) in the mixture?

Knowledge Check

  • A gas mixture was prepared by taking equal mole of CO and N_(2) . If total pressure of the mixture was 1 atm, the partial pressure of N_(2) in the mixture is-

    A
    0.5 atm
    B
    0.8 atm
    C
    0.9 atm
    D
    1 atm

  • A gas mixture consisting of 1 mol of N_(2) and 3 mol of O_(2) has a pressure of 2 atmm at 0^(@)C . Keeping the volume and the temperature of the mixture constant, some amount of O_(2) was removed from the mixture. As a result, the total pressure of the mixture and the partial pressure of N_(2) in the mixture became 1.5 atm and 0.5 atm respectively. the amount of oxygen gas removed was-

    A
    8 g
    B
    16 g
    C
    32 g
    D
    64 g

    • Similar Questions

    Explore conceptually related problems

    Calculate the partial pressure of SO_2 in a mixture with 60% CO_2 and total pressure being 4 atm.

    A mixture of N_(2) and O_(2) at 1 bar pressure contains 80% N_(2) by weight. Calculate the partial pressure of N_(2) in the mixture.

    In a 10 litre volumetric flask contains 1 gram He and 6.4 and gram O_2 at 27^@C temperature If that total pressure of the mixture is 1.107 atmosphere then what is the partial pressure of He and O_2 ?

    A gaseous mixture was prepared by taking equal moles of CO and N_2 . If the total pressure of the mixture was found 1 atm, the partial pressure of the nitrogen (N_2) in the mixture is

    At a given temperature, NH_3 gas is in a state of equilibrium with N_2 and H_2 gases in a closed container. If the partial pressures of N_2 and H_2 in the mixture are 1.34 and 0.67 atm respectively and the total pressure of the reaction mixture is 4.67 atm, then determine the value of K_p for the reaction : N_2(g)+3H_2(g)

    At 27^(@)C , a cylinder of volume 10 L contains a gas mixture consisting of 0.4 He, 1.6g O_(2)&1.4g H_(2) . Determine the total pressure of the mixture and the partial pressure of He in the mixture.

    Home
    Profile