The pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. - Chemistry

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Numerical

The pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

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Solution

For ideal gas A, the ideal gas equation is given by,

`"P"_"A""V" = "n"_"A""RT"`  ............(i)

Where, pA and nA represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

`"p"_"s""V" = "n"_"s""RT"`  ......(ii)

Where, pand nB represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

From equation (i), we have

`"p"_"AV" = "m"_"A"/"M"_"A" "RT" => ("p"_"A""M"_"A")/"m"_"A"  = ("RT")/"V"`   .....(iii)

From equation (ii), we have

`"p"_"BV" = "m"_"B"/"M"_"B"  "RT" => ("p"_"B""M"_"B")/"m"_"B" = "RT"/"V"` .......(iv)

Where, Mand MB are the molecular masses of gases A and B respectively.

Now, from equations (iii) and (iv), we have

`("p"_"A""M"_"A")/"m"_"A" = ("p"_"B""M"_"B")/"m"_"B"`  .....(v)

Given

`"m"_"A" = 1 "g"`

`"p"_"A" =  2 "bar"`

`"m"_"B" = 2"g"`

`"p"_"B" = (3 - 2) = 1 "bar"`

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

`(2xx"M"_"A")/1 = (1xx"M"_"B")/2`

`=> 4"M"_"A" = "M"_"B"`

Thus, a relationship between the molecular masses of A and B is given by

`4"M"_"A" = "M"_"B"`

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Chapter 5: States of Matter - EXERCISES [Page 158]

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NCERT Chemistry Part 1 and 2 Class 11
Chapter 5 States of Matter
EXERCISES | Q 5.5 | Page 158

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