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A 50 litre vessel is equally divided int...

A 50 litre vessel is equally divided into three parts with the help of two stationary semi permeable membrane ( SPM). The vessel contains 60g `H_(2)` gas in the left chambe, 160g `O_(2)` in the middle & 140 g `N_(2)` in the right one . The left SPM allows transfer of only `H_(2) `gas while the right one allows the transfer of both `H_(2) `& `N_(2)`. Calculate the final ratio of pressure in the three chambers.

Text Solution

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The correct Answer is:
`4:7 : 5`
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Knowledge Check

  • A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left , middle and right parts are filled with m_(H) = 30 gm of hydrogen , m_(O_(2)) = 160 gm of oxygen and m_(N) = 70 gm of nitrogen respectively . The left partition lets through only hydrogen while the right partition lets through hydrogen and nitrogen . If the temperature in all is 300K , the ratio of pressure in the three compartments will be :

    A
    `4 : 9: 5`
    B
    `1.3 : 4.5 : 2`
    C
    `9 : 4 : 5`
    D
    `9: 5: 4`

  • A gaseous mixture contains 1 g of H_(2) , 4 g of He , 7 g of N_(2) and 8 g of O_(2) . The gas having the highest partial pressure is :-

    A
    `H_(2)`
    B
    `O_(2)`
    C
    He
    D
    `N_(2)`

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