Question

initially in a container 1 g of gas a has 4 atm pressure at constant temperature if 2 g of gas b is addded in same container at same temperature then pressure becomes 6 atm what will be the ratio of molecular weight of a and b

Answer 1

applying formula,

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

where n1 is initial number of moles of gas, n2 is final number of moles of gas ,P1 is initial pressure, P2 is final pressure , V1 is initial volume and V2 is final volume of gas.

given, P1 = 4atm, P2 = 6atm , V1 = V2 = V(let)

number of moles of gas a , n1 = 1g/Ma

number of moles of gas b, n = 2g/Mb

so, n2 = n1 + n = 1/Ma + 2/Mb

now, 4V/{1/Ma } = 6V/{1/Ma + 2/Mb}

or, 2Ma = {3Ma.Mb}/{Mb + 2Ma}

or, 2Ma.Mb + 4Ma² = 3Ma.Mb

or, 4Ma² = Ma.Mb

or, Ma/Mb = 1/4

hence, ratio of molecular weight of gas a to gas b is 1 : 4

Answer 2

Answer:

Mb=4Ma

Explanation:

We have the formula that is

P1V1/n1=P2V2/n2

where P1 is the initial pressure of the gas A and P2 is the pressure of the gas b and V1 and V2 are the volumes which is equal to V and n1 and n2 is the number of moles of a and b gas respectively

Then according to Dalton law total pressure is the sum of its individual pressure

Total pressure is equal to 6

P2=6-4=2

n1=1g/Ma,n2=2g/Mb

Now according to formula

P1/P2=n1/n2

4/2=1/Ma/2/Mb

2=Mb/2*Ma

Mb/Ma=4

Mb=4Ma