Chemistry, asked by usman4144, 11 months ago

initially in a container 1gm of gas A has 4 atm pressure at constant temperature.if 2gm of gas B is added in same cotainer at same temperature then pressure becomes 6 atm ,what will be the ratio of molecular weight of A and B ​

Answers

Answered by BarrettArcher
9

Answer : The ratio of molecular weight of A and B ​is, 3 : 4

Explanation :

According to the ideal gas equation,

PV=nRT

At constant volume of temperature,

P=n

or,

P=\frac{w}{M}

The expression used for to gases will be,

M=\frac{w}{P}

\frac{M_A}{M_B}=\frac{w_A\times P_B}{w_B\times P_A}   .......(1)

where,

P_A = pressure of gas A = 4 atm

P_B = pressure of gas B= 6 atm

w_A = mass of gas A = 1 g

w_B = mass of gas B= 2 g

M_A = molecular weight of gas A

M_B = molecular weight of gas B

Now put all the given values in equation 1, we get the ratio of molecular weight of A and B.

\frac{M_A}{M_B}=\frac{1g\times 6atm}{2g\times 4atm}

\frac{M_A}{M_B}=\frac{3}{4}

Therefore, the ratio of molecular weight of A and B ​is, 3 : 4

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