Q.1 vanderwaal equation for CH4 at low pressure is

a)PV=RT-Pb b)PV=RT-a/V c)PV=RT+a/V D)PV=RT+Pb

Ans=b

Q.2 when 0.75 mol of solid A4 and 2 mol O2(g) are heated in a sealed vessel completely using up the reactants and producing only 1 compound. it is found that when the temp is reduced to the initial temp the content of the vessel exhibit a pressure equal to half the original pressure. the molecular formula of the product would be

Ans=A3O4

Q.3 A mix. of HCOOH AND (COOH)2 is heated with conc. H2SO4. the gas produced is collected and on its treatment with KOH soln the vol of gas dec. by 1/6th. the molar ratio of 2 acids in the original mix. is

ans=4:1

Q.4 A mix of H2 and He in 1:1 mass ratio is allowed to diffuse through a porous pot. The mol ratio of H2 to He in the initially effused mix is

ans=2√2:1

Ans .2 > (x* represents x dash for any variable x)

for gas eqtn (initial conditio) PV=nRT ________(1)

n=  2 (of O2) (3/4 moles of A wuldnt be incluided becoz its solid nd only gases are to be included.)

hence n=2

nd for final condition (after reducing the temp to initial)

P*v*=n*RT* __________(2)

now from questn.. T* = T, P* = P/2 , V=v*(container is seald)

hence by (1)/(2)

n/n* = 2  ....  n*=1

now if we write reaction as 3/4 A4 + 2O2 --> 1(let x)

that x shuld include all the 3 atoms of A nd 4 atoms of O (on LHS of reaction).

hence ans .. A3O4

Approved

ans 1.

Vanderwalls eqtn (origoinal) - (P + a/V² ) (V-b)=RT ------- (1)

when pressure is low, volume is high (as per boyles law).

hence (V-b) can be written as (V) coz b wont affect it much.

so replace (V-b) by (V) in eqtn 1.

ull get ans B.

for Q1

if V inc then V^2 increases very much then a/v^2 decreases so why is a/V^2 not neglected?

you keep in mind the point that we ignore a quantity saying its too small only in addition or substraction as in this case.

we cant ignore a/V^2 though its small coz here P is also small as per the given question ... hence it will afect the value of P.

Consider if .00001+ 10 here 1*10^-5 wont effect much the ans of this statement.

whille if we do 0.001 + 0.00001 = we cant ignor any 1 quantity coz it will surely affect the ans.