If masses of all molecule of a gas are halved and their speed doubled then the ratio of initial and final pressure will be

Explanation

Given, `m_(1)=m,v_(1)=v,V_1=V,m_(2)-m//2,v_(2)=2v,V_(2)=V_(1)=V`
`therefore` pressure exerted by first gas molecule `P_(1)=(1)/(3)m_(1)/(V_(1))v_(1)"^(2)`
and pressure exerted by second gas molecule
`P_(2)=(1)/(3)m_(2)/(V_(2))v_(2)"^(2)`
`therefore" " P_(1)/(P_(2))=m_(1)/m_(2)xxV_(2)/V_(1)xx(v_(1)"^(2))/(v_(2)^(2))`
Putting the values, we get `P_(1)/P_(2)=(m)/(m//2)xx(V)/(V)xx((v)/(2v))^(2)=2xx(1)/(4)=(1)/(2)`