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Nagma Kata
Asked: 2022-11-08T10:18:37+05:30 In:

Containers `A` and `B` have same, gases. Pressure, volume and temperature of `A` are all twice that of `B`, then the ratio of number of molecules of `A` and `B` are
A. `1 : 2`
B. `2 : 1`
C. `1 : 4`
D. `4 : 1`

Containers `A` and `B` have same, gases. Pressure, volume and temperature of `A` are all twice that of `B`, then the ratio of number of molecules of `A` and `B` are
A. `1 : 2`
B. `2 : 1`
C. `1 : 4`
D. `4 : 1`
Parabola
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  1. Correct Answer – B
    Applying the ideal gas equation , we have for container `A`,
    `p_(A)V_(A) = n_(A)RT_(A)`
    For container `B , p_(B)V_(B) = n_(B)RT_(B)`
    Dividing the two , we get
    `(p_(A)V_(A))/(p_(B)V_(B)) = (n_(A)RT_(A))/(n_(B)RT_(B))`
    Since `p_(A) = 2p_(B) , V_(A) = 2 V_(B)` , and `T_(A) = 2 T_(B)` , we have
    `((2p_(B))(2V_(B)))/(p_(B)V_(B)) = (n_(A))/(n_(B)) = ((2T_(B)))/((T_(B)))`
    ` 4 = (n_(A))/(n_(B)) = 2`
    or `(n_(A))/(n_(B)) = (4)/(2) = (2)/(1)`

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Munni Ram Gopal Vyas
Asked: 2022-10-29T01:55:42+05:30 In:

Containers `A` and `B` have same, gases. Pressure, volume and temperature of `A` are all twice that of `B`, then the ratio of number of molecules of `A` and `B` are

Containers `A` and `B` have same, gases. Pressure, volume and temperature of `A` are all twice that of `B`, then the ratio of number of molecules of `A` and `B` are
Parabola
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1 Answer

  1. Correct Answer – B
    According to gas law
    `PV= nRT, n=(PV)/(RT)`
    `(n_(A))/(n_(B))=((P_(1)V_(1))/(RT_(1)))/((P_(2)V_(2))/(RT_(2))),(n_(A))/(n_(B))=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2)V_(2))`
    `(n_(A))/(n_(B))=(2Pxx2V)/(2T)xx(T)/(PV), (n_A)/(n_(B))=(2)/(1)`

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