The correct option is D C4H10
CnH2n+2+(3n+12)O2→nCO2+(n+1)H2O
Let initial pressure of CnH2n+2 is P, then increase in pressure :
=P[(2n+1)−1−(3n+12)]
=(n−12)P
For a constant volume of the steel vessel, T∝P
So,
At 546 K pressure is 4.6 atm
∴ at 5462=273 K pressure is 4.62=2.3 atm;
Increase in pressure= (2.3−2) atm=0.3 atmR=ideal gas constant=0.0821 L.atm mol−1 K−1V=volume of steel vessel=22.41 L
P=nRTV=11.6M×(0.0821×27322.41)
P=11.6M
where M = Molecular weight of organic compound
M=n×12+(2n+2)×1M=14n+2
Since,
(n−12)P=0.3
Putting P=11.614n+2
⇒(n−12)×11.6(14n+2)=0.3
⇒n−114n+2=0.611.6
⇒n=4
∴ Compound is C4H10