# 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.

#### 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"

Is there an error in this question or solution?
Chapter 5: States of Matter - EXERCISES [Page 158]

#### APPEARS IN

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

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